Chapter 4 from Textbook Rs Aggarwal 2021 2022 for Class 10 MATHS

Question 1:

Answer:

$i) (x^{2} - x + 3) is a quadratic polynomial . ∴ x^{2} - x + 3 = 0 is a quadratic equation . ii) C learly, (2 x^{2} + \frac{5}{2} x - \sqrt{3}) is a quadratic polynomial . ∴ 2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0 is a quadratic equation . iii) C learly, (\sqrt{2} x^{2} + 7 x + 5 \sqrt{2}) is a quadratic polynomial . ∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 is a quadratic equation . iv) C learly, (\frac{1}{3} x^{2} + \frac{1}{5} x - 2) is a quadratic polynomial . ∴ \frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0 is a quadratic equation . v) (x^{2} - 3 x - \sqrt{x} + 4) contains a term with \sqrt{x}, i .e, x^{\frac{1}{2}}, where \frac{1}{2} is not a integer . Therefore, it is not a quadratic polynomial . ∴ x^{2} - 3 x - \sqrt{x} + 4 = 0 is not a quadratic equation . vi) x - \frac{6}{x} = 3 \Rightarrow x^{2} - 6 = 3 x \Rightarrow x^{2} - 3 x - 6 = 0 \begin{array}{l} (x^{2} - 3 x - 6) is a quadratic polynomial; therefore, the given \\ equation is quadratic . \end{array} vii) x + \frac{2}{x} = x^{2} \Rightarrow x^{2} + 2 = x^{3} \Rightarrow x^{3} - x^{2} - 2 = 0 (x^{3} - x^{2} - 2) is not a quadratic polynomial . ∴ x^{3} - x^{2} - 2 = 0 is not a quadratic equation . viii) x^{2} - \frac{1}{x^{2}} = 5 \Rightarrow x^{4} - 1 = 5 x^{2} \Rightarrow x^{4} - 5 x^{2} - 1 = 0 (x^{4} - 5 x^{2} - 1) is a polynomial with degree 4 . ∴ x^{4} - 5 x^{2} - 1 = 0 is not a quadratic equation .$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
$\Rightarrow x^{3} + 6 x^{2} + 12 x + 8 = x^{3} - 8 \Rightarrow 6 x^{2} + 12 x + 16 = 0$
This is of the form ax²+ bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
$\Rightarrow 6 x^{2} + 4 x + 9 x + 6 = 6 (x^{2} - 3 x + 2) \Rightarrow 6 x^{2} + 13 x + 6 = 6 x^{2} - 18 x + 12 \Rightarrow 31 x - 6 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$
$\Rightarrow {(\frac{x^{2} + 1}{x})}^{2} = 2 (\frac{x^{2} + 1}{x}) + 3 \Rightarrow {(x^{2} + 1)}^{2} = 2 x (x^{2} + 1) + 3 x^{2} \Rightarrow x^{4} + 2 x^{2} + 1 = 2 x^{3} + 2 x + 3 x^{2} \Rightarrow x^{4} - 2 x^{3} - x^{2} - 2 x + 1 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.

Question 2:

$i) (x^{2} - x + 3) is a quadratic polynomial . ∴ x^{2} - x + 3 = 0 is a quadratic equation . ii) C learly, (2 x^{2} + \frac{5}{2} x - \sqrt{3}) is a quadratic polynomial . ∴ 2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0 is a quadratic equation . iii) C learly, (\sqrt{2} x^{2} + 7 x + 5 \sqrt{2}) is a quadratic polynomial . ∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 is a quadratic equation . iv) C learly, (\frac{1}{3} x^{2} + \frac{1}{5} x - 2) is a quadratic polynomial . ∴ \frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0 is a quadratic equation . v) (x^{2} - 3 x - \sqrt{x} + 4) contains a term with \sqrt{x}, i .e, x^{\frac{1}{2}}, where \frac{1}{2} is not a integer . Therefore, it is not a quadratic polynomial . ∴ x^{2} - 3 x - \sqrt{x} + 4 = 0 is not a quadratic equation . vi) x - \frac{6}{x} = 3 \Rightarrow x^{2} - 6 = 3 x \Rightarrow x^{2} - 3 x - 6 = 0 \begin{array}{l} (x^{2} - 3 x - 6) is a quadratic polynomial; therefore, the given \\ equation is quadratic . \end{array} vii) x + \frac{2}{x} = x^{2} \Rightarrow x^{2} + 2 = x^{3} \Rightarrow x^{3} - x^{2} - 2 = 0 (x^{3} - x^{2} - 2) is not a quadratic polynomial . ∴ x^{3} - x^{2} - 2 = 0 is not a quadratic equation . viii) x^{2} - \frac{1}{x^{2}} = 5 \Rightarrow x^{4} - 1 = 5 x^{2} \Rightarrow x^{4} - 5 x^{2} - 1 = 0 (x^{4} - 5 x^{2} - 1) is a polynomial with degree 4 . ∴ x^{4} - 5 x^{2} - 1 = 0 is not a quadratic equation .$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
$\Rightarrow x^{3} + 6 x^{2} + 12 x + 8 = x^{3} - 8 \Rightarrow 6 x^{2} + 12 x + 16 = 0$
This is of the form ax²+ bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
$\Rightarrow 6 x^{2} + 4 x + 9 x + 6 = 6 (x^{2} - 3 x + 2) \Rightarrow 6 x^{2} + 13 x + 6 = 6 x^{2} - 18 x + 12 \Rightarrow 31 x - 6 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$
$\Rightarrow {(\frac{x^{2} + 1}{x})}^{2} = 2 (\frac{x^{2} + 1}{x}) + 3 \Rightarrow {(x^{2} + 1)}^{2} = 2 x (x^{2} + 1) + 3 x^{2} \Rightarrow x^{4} + 2 x^{2} + 1 = 2 x^{3} + 2 x + 3 x^{2} \Rightarrow x^{4} - 2 x^{3} - x^{2} - 2 x + 1 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.

Answer:

${The given equation is (3x}^{2} + 2 x - 1 = 0) . (i) x = (- 1) L .H .S . = x^{2} + 2 x - 1 = 3 \times {(- 1)}^{2} + 2 \times (- 1) - 1 = 3 - 2 - 1 = 0 = R .H .S . Thus, (- 1) is a root of {(3x}^{2} + 2 x - 1 = 0) . (ii) On substituting x = \frac{1}{3} in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(\frac{1}{3})}^{2} + 2 \times \frac{1}{3} - 1 = 3^{1} \times \frac{1}{9_{3}} + \frac{2}{3} - 1 = \frac{1 + 2 - 3}{3} = \frac{0}{3} = 0 = R .H .S . Thus, (\frac{1}{3}) is a root of {(3x}^{2} + 2 x - 1 = 0) . (iii) On substituting x = (- \frac{1}{2}) in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(- \frac{1}{2})}^{2} + 2^{1} \times (- \frac{1}{2_{1}}) - 1 = 3 \times \frac{1}{4} - 1 - 1 = \frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{- 5}{4} \neq 0 Thus, L .H .S . = R .H .S . Hence, (- \frac{1}{2}) is a not solution of (3 x^{2} + 2 x - 1 = 0) .$

Question 3:

${The given equation is (3x}^{2} + 2 x - 1 = 0) . (i) x = (- 1) L .H .S . = x^{2} + 2 x - 1 = 3 \times {(- 1)}^{2} + 2 \times (- 1) - 1 = 3 - 2 - 1 = 0 = R .H .S . Thus, (- 1) is a root of {(3x}^{2} + 2 x - 1 = 0) . (ii) On substituting x = \frac{1}{3} in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(\frac{1}{3})}^{2} + 2 \times \frac{1}{3} - 1 = 3^{1} \times \frac{1}{9_{3}} + \frac{2}{3} - 1 = \frac{1 + 2 - 3}{3} = \frac{0}{3} = 0 = R .H .S . Thus, (\frac{1}{3}) is a root of {(3x}^{2} + 2 x - 1 = 0) . (iii) On substituting x = (- \frac{1}{2}) in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(- \frac{1}{2})}^{2} + 2^{1} \times (- \frac{1}{2_{1}}) - 1 = 3 \times \frac{1}{4} - 1 - 1 = \frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{- 5}{4} \neq 0 Thus, L .H .S . = R .H .S . Hence, (- \frac{1}{2}) is a not solution of (3 x^{2} + 2 x - 1 = 0) .$

Answer:

(i)
$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$
So, the equation becomes $x^{2} - 4 x + 3 = 0$
On factorising we get;
$x^{2} - x - 3 x + 3 = 0 x (x - 1) - 3 (x - 1) = 0 (x - 1) (x - 3) = 0 \Rightarrow x - 1 = 0 or x - 3 = 0 \Rightarrow x = 1 or x = 3$
Hence, the other root is 3.

(ii)
$It is given that \frac{3}{4} is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(\frac{3}{4})}^{2} + b \times \frac{3}{4} - 6 = 0 \Rightarrow \frac{9 a}{16} + \frac{3 b}{4} = 6 \Rightarrow \frac{9 a + 12 b}{16} = 6 \Rightarrow 9 a + 12 b - 96 = 0 \Rightarrow 3 a + 4 b = 32 ... (i) Again, (- 2) is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(- 2)}^{2} + b \times (- 2) - 6 = 0 \Rightarrow 4 a - 2 b = 6 \Rightarrow 2 a - b = 3 ... (ii) On multiplying (ii) by 4 and adding the result with (i), w e get: \Rightarrow 3 a + 4 b + 8 a - 4 b = 32 + 12 \Rightarrow 11 a = 44 \Rightarrow a = 4 Putting the value of a in (ii), we get : 2 \times 4 - b = 3 \Rightarrow 8 - b = 3 \Rightarrow b = 5 Hence, the required values of a and b are 4 and 5, respectively.$

Question 4:

(i)
$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$
So, the equation becomes $x^{2} - 4 x + 3 = 0$
On factorising we get;
$x^{2} - x - 3 x + 3 = 0 x (x - 1) - 3 (x - 1) = 0 (x - 1) (x - 3) = 0 \Rightarrow x - 1 = 0 or x - 3 = 0 \Rightarrow x = 1 or x = 3$
Hence, the other root is 3.

(ii)
$It is given that \frac{3}{4} is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(\frac{3}{4})}^{2} + b \times \frac{3}{4} - 6 = 0 \Rightarrow \frac{9 a}{16} + \frac{3 b}{4} = 6 \Rightarrow \frac{9 a + 12 b}{16} = 6 \Rightarrow 9 a + 12 b - 96 = 0 \Rightarrow 3 a + 4 b = 32 ... (i) Again, (- 2) is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(- 2)}^{2} + b \times (- 2) - 6 = 0 \Rightarrow 4 a - 2 b = 6 \Rightarrow 2 a - b = 3 ... (ii) On multiplying (ii) by 4 and adding the result with (i), w e get: \Rightarrow 3 a + 4 b + 8 a - 4 b = 32 + 12 \Rightarrow 11 a = 44 \Rightarrow a = 4 Putting the value of a in (ii), we get : 2 \times 4 - b = 3 \Rightarrow 8 - b = 3 \Rightarrow b = 5 Hence, the required values of a and b are 4 and 5, respectively.$

Answer:

LHS;
Consider the quadratic equation;
$a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0$
Put $x = - \frac{b c}{a d}$ in the given equation.
$a d^{2} [\frac{a (- \frac{b c}{a d})}{b} + \frac{2 c}{d}] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [a d (- \frac{b c}{a d}) + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [- b c + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} (b c) (- \frac{b c}{a d}) + b c^{2} = - \frac{a b^{2} c^{2} d^{2}}{a b d^{2}} + b c^{2} = - b c^{2} + b c^{2} = 0 = RHS$
Hence, $x = - \frac{b c}{a d}$ is a solution to the given quadratic equation.

Question 5:

LHS;
Consider the quadratic equation;
$a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0$
Put $x = - \frac{b c}{a d}$ in the given equation.
$a d^{2} [\frac{a (- \frac{b c}{a d})}{b} + \frac{2 c}{d}] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [a d (- \frac{b c}{a d}) + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [- b c + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} (b c) (- \frac{b c}{a d}) + b c^{2} = - \frac{a b^{2} c^{2} d^{2}}{a b d^{2}} + b c^{2} = - b c^{2} + b c^{2} = 0 = RHS$
Hence, $x = - \frac{b c}{a d}$ is a solution to the given quadratic equation.

Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

⇒ x = $\frac{3}{2}$ or x = $- \frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $- \frac{1}{3}$ .

Question 6:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

⇒ x = $\frac{3}{2}$ or x = $- \frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $- \frac{1}{3}$ .

Answer:

4x² + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = $- \frac{5}{4}$

Hence, the roots of the given equation are 0 and $- \frac{5}{4}$ .

Question 7:

4x² + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = $- \frac{5}{4}$

Hence, the roots of the given equation are 0 and $- \frac{5}{4}$ .

Answer:

$Given: 3 x^{2} - 243 = 0 \Rightarrow 3 (x^{2} - 81) = 0 \Rightarrow {(x)}^{2} - {(9)}^{2} = 0 \Rightarrow (x + 9) (x - 9) = 0 \Rightarrow x + 9 = 0 or x - 9 = 0 \Rightarrow x = - 9 or x = 9 Hence, - 9 and 9 are the roots of the equation 3 x^{2} - 243 = 0.$

Question 8:

$Given: 3 x^{2} - 243 = 0 \Rightarrow 3 (x^{2} - 81) = 0 \Rightarrow {(x)}^{2} - {(9)}^{2} = 0 \Rightarrow (x + 9) (x - 9) = 0 \Rightarrow x + 9 = 0 or x - 9 = 0 \Rightarrow x = - 9 or x = 9 Hence, - 9 and 9 are the roots of the equation 3 x^{2} - 243 = 0.$

Answer:

We write, $x = 4 x - 3 x$ as $2 x^{2} \times (- 6) = - 12 x^{2} = 4 x \times (- 3 x)$

$∴ 2 x^{2} + x - 6 = 0 \Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0 \Rightarrow 2 x (x + 2) - 3 (x + 2) = 0 \Rightarrow (x + 2) (2 x - 3) = 0$

$\Rightarrow x + 2 = 0 or 2 x - 3 = 0 \Rightarrow x = - 2 or x = \frac{3}{2}$
Hence, the roots of the given equation are $- 2$ and $\frac{3}{2}$ .

Question 9:

We write, $x = 4 x - 3 x$ as $2 x^{2} \times (- 6) = - 12 x^{2} = 4 x \times (- 3 x)$

$∴ 2 x^{2} + x - 6 = 0 \Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0 \Rightarrow 2 x (x + 2) - 3 (x + 2) = 0 \Rightarrow (x + 2) (2 x - 3) = 0$

$\Rightarrow x + 2 = 0 or 2 x - 3 = 0 \Rightarrow x = - 2 or x = \frac{3}{2}$
Hence, the roots of the given equation are $- 2$ and $\frac{3}{2}$ .

Answer:

We write, $6 x = x + 5 x$ as $x^{2} \times 5 = 5 x^{2} = x \times 5 x$

$∴ x^{2} + 6 x + 5 = 0 \Rightarrow x^{2} + x + 5 x + 5 = 0 \Rightarrow x (x + 1) + 5 (x + 1) = 0 \Rightarrow (x + 1) (x + 5) = 0$

$\Rightarrow x + 1 = 0 or x + 5 = 0 \Rightarrow x = - 1 or x = - 5$

Hence, the roots of the given equation are −1 and −5.

Question 10:

We write, $6 x = x + 5 x$ as $x^{2} \times 5 = 5 x^{2} = x \times 5 x$

$∴ x^{2} + 6 x + 5 = 0 \Rightarrow x^{2} + x + 5 x + 5 = 0 \Rightarrow x (x + 1) + 5 (x + 1) = 0 \Rightarrow (x + 1) (x + 5) = 0$

$\Rightarrow x + 1 = 0 or x + 5 = 0 \Rightarrow x = - 1 or x = - 5$

Hence, the roots of the given equation are −1 and −5.

Answer:

We write, $- 3 x = 3 x - 6 x$ as $9 x^{2} \times (- 2) = - 18 x^{2} = 3 x \times (- 6 x)$

$∴ 9 x^{2} - 3 x - 2 = 0 \Rightarrow 9 x^{2} + 3 x - 6 x - 2 = 0 \Rightarrow 3 x (3 x + 1) - 2 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x - 2) = 0$
$\Rightarrow 3 x + 1 = 0 or 3 x - 2 = 0 \Rightarrow x = - \frac{1}{3} or x = \frac{2}{3}$
Hence, the roots of the given equation are $- \frac{1}{3}$ and $\frac{2}{3}$ .

Question 11:

We write, $- 3 x = 3 x - 6 x$ as $9 x^{2} \times (- 2) = - 18 x^{2} = 3 x \times (- 6 x)$

$∴ 9 x^{2} - 3 x - 2 = 0 \Rightarrow 9 x^{2} + 3 x - 6 x - 2 = 0 \Rightarrow 3 x (3 x + 1) - 2 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x - 2) = 0$
$\Rightarrow 3 x + 1 = 0 or 3 x - 2 = 0 \Rightarrow x = - \frac{1}{3} or x = \frac{2}{3}$
Hence, the roots of the given equation are $- \frac{1}{3}$ and $\frac{2}{3}$ .

Answer:

$Given : x^{2} + 12 x + 35 = 0 \Rightarrow x^{2} + 7 x + 5 x + 35 = 0 \Rightarrow x (x + 7) + 5 (x + 7) = 0 \Rightarrow (x + 5) (x + 7) = 0 \Rightarrow x + 5 = 0 or x + 7 = 0 \Rightarrow x = - 5 or x = - 7 Hence, - 5 and - 7 are the roots of the equation x^{2} + 12 x + 35 = 0.$

Question 12:

$Given : x^{2} + 12 x + 35 = 0 \Rightarrow x^{2} + 7 x + 5 x + 35 = 0 \Rightarrow x (x + 7) + 5 (x + 7) = 0 \Rightarrow (x + 5) (x + 7) = 0 \Rightarrow x + 5 = 0 or x + 7 = 0 \Rightarrow x = - 5 or x = - 7 Hence, - 5 and - 7 are the roots of the equation x^{2} + 12 x + 35 = 0.$

Answer:

$Given : x^{2} = 18 x - 77 \Rightarrow x^{2} - 18 x + 77 = 0 \Rightarrow x^{2} - (11 x + 7 x) + 77 = 0 \Rightarrow x^{2} - 11 x - 7 x + 77 = 0 \Rightarrow x (x - 11) - 7 (x - 11) = 0 \Rightarrow (x - 7) (x - 11) = 0 \Rightarrow x - 7 = 0 or x - 11 = 0 \Rightarrow x = 7 or x = 11 Hence, 7 and 11 are the roots of the equation x^{2} = 18 x - 77.$

Question 13:

$Given : x^{2} = 18 x - 77 \Rightarrow x^{2} - 18 x + 77 = 0 \Rightarrow x^{2} - (11 x + 7 x) + 77 = 0 \Rightarrow x^{2} - 11 x - 7 x + 77 = 0 \Rightarrow x (x - 11) - 7 (x - 11) = 0 \Rightarrow (x - 7) (x - 11) = 0 \Rightarrow x - 7 = 0 or x - 11 = 0 \Rightarrow x = 7 or x = 11 Hence, 7 and 11 are the roots of the equation x^{2} = 18 x - 77.$

Answer:

$Given : 6 x^{2} + 11 x + 3 = 0 \Rightarrow 6 x^{2} + 9 x + 2 x + 3 = 0 \Rightarrow 3 x (2 x + 3) + 1 (2 x + 3) = 0 \Rightarrow (3 x + 1) (2 x + 3) = 0 \Rightarrow 3 x + 1 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 3}{2} Hence, \frac{- 1}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + 11 x + 3 = 0.$

Question 14:

$Given : 6 x^{2} + 11 x + 3 = 0 \Rightarrow 6 x^{2} + 9 x + 2 x + 3 = 0 \Rightarrow 3 x (2 x + 3) + 1 (2 x + 3) = 0 \Rightarrow (3 x + 1) (2 x + 3) = 0 \Rightarrow 3 x + 1 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 3}{2} Hence, \frac{- 1}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + 11 x + 3 = 0.$

Answer:

$Given: 6 x^{2} + x - 12 = 0 \Rightarrow 6 x^{2} + 9 x - 8 x - 12 = 0 \Rightarrow 3 x (2 x + 3) - 4 (2 x + 3) = 0 \Rightarrow (3 x - 4) (2 x + 3) = 0 \Rightarrow 3 x - 4 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{4}{3} or x = \frac{- 3}{2} Hence, \frac{4}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + x - 12 = 0.$

Question 15:

$Given: 6 x^{2} + x - 12 = 0 \Rightarrow 6 x^{2} + 9 x - 8 x - 12 = 0 \Rightarrow 3 x (2 x + 3) - 4 (2 x + 3) = 0 \Rightarrow (3 x - 4) (2 x + 3) = 0 \Rightarrow 3 x - 4 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{4}{3} or x = \frac{- 3}{2} Hence, \frac{4}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + x - 12 = 0.$

Answer:

We write, $- 2 x = - 3 x + x$ as $3 x^{2} \times (- 1) = - 3 x^{2} = (- 3 x) \times x$

$∴ 3 x^{2} - 2 x - 1 = 0 \Rightarrow 3 x^{2} - 3 x + x - 1 = 0 \Rightarrow 3 x (x - 1) + 1 (x - 1) = 0 \Rightarrow (x - 1) (3 x + 1) = 0$
$\Rightarrow x - 1 = 0 or 3 x + 1 = 0 \Rightarrow x = 1 or x = - \frac{1}{3}$
Hence, the roots of the given equation are $1$ and $- \frac{1}{3}$ .

Question 16:

We write, $- 2 x = - 3 x + x$ as $3 x^{2} \times (- 1) = - 3 x^{2} = (- 3 x) \times x$

$∴ 3 x^{2} - 2 x - 1 = 0 \Rightarrow 3 x^{2} - 3 x + x - 1 = 0 \Rightarrow 3 x (x - 1) + 1 (x - 1) = 0 \Rightarrow (x - 1) (3 x + 1) = 0$
$\Rightarrow x - 1 = 0 or 3 x + 1 = 0 \Rightarrow x = 1 or x = - \frac{1}{3}$
Hence, the roots of the given equation are $1$ and $- \frac{1}{3}$ .

Answer:

$Given: 4 x^{2} - 9 x = 100 \Rightarrow 4 x^{2} - 9 x - 100 = 0 \Rightarrow 4 x^{2} - (25 x - 16 x) - 100 = 0 \Rightarrow 4 x^{2} - 25 x + 16 x - 100 = 0 \Rightarrow x (4 x - 25) + 4 (4 x - 25) = 0 \Rightarrow (4 x - 25) (x + 4) = 0 \Rightarrow 4 x - 25 = 0 or x + 4 = 0 \Rightarrow x = \frac{25}{4} or x = - 4 Hence, the roots of the equation are \frac{25}{4} and - 4 .$

Question 17:

$Given: 4 x^{2} - 9 x = 100 \Rightarrow 4 x^{2} - 9 x - 100 = 0 \Rightarrow 4 x^{2} - (25 x - 16 x) - 100 = 0 \Rightarrow 4 x^{2} - 25 x + 16 x - 100 = 0 \Rightarrow x (4 x - 25) + 4 (4 x - 25) = 0 \Rightarrow (4 x - 25) (x + 4) = 0 \Rightarrow 4 x - 25 = 0 or x + 4 = 0 \Rightarrow x = \frac{25}{4} or x = - 4 Hence, the roots of the equation are \frac{25}{4} and - 4 .$

Answer:

$Given: 15 x^{2} - 28 = x \Rightarrow 15 x^{2} - x - 28 = 0 \Rightarrow 15 x^{2} - (21 x - 20 x) - 28 = 0 \Rightarrow 15 x^{2} - 21 x + 20 x - 28 = 0 \Rightarrow 3 x (5 x - 7) + 4 (5 x - 7) = 0 \Rightarrow (3 x + 4) (5 x - 7) = 0 \Rightarrow 3 x + 4 = 0 or 5 x - 7 = 0 \Rightarrow x = \frac{- 4}{3} or x = \frac{7}{5} H ence, the roots of the equation are \frac{- 4}{3} and \frac{7}{5} .$

Question 18:

$Given: 15 x^{2} - 28 = x \Rightarrow 15 x^{2} - x - 28 = 0 \Rightarrow 15 x^{2} - (21 x - 20 x) - 28 = 0 \Rightarrow 15 x^{2} - 21 x + 20 x - 28 = 0 \Rightarrow 3 x (5 x - 7) + 4 (5 x - 7) = 0 \Rightarrow (3 x + 4) (5 x - 7) = 0 \Rightarrow 3 x + 4 = 0 or 5 x - 7 = 0 \Rightarrow x = \frac{- 4}{3} or x = \frac{7}{5} H ence, the roots of the equation are \frac{- 4}{3} and \frac{7}{5} .$

Answer:

$Given: 4 - 11 x = 3 x^{2} \Rightarrow 3 x^{2} + 11 x - 4 = 0 \Rightarrow 3 x^{2} + 12 x - x - 4 = 0 \Rightarrow 3 x (x + 4) - 1 (x + 4) = 0 \Rightarrow (x + 4) (3 x - 1) = 0 \Rightarrow x + 4 = 0 or 3 x - 1 = 0 \Rightarrow x = - 4 or x = \frac{1}{3} H ence, the roots of the equation are - 4 and \frac{1}{3} .$

Question 19:

$Given: 4 - 11 x = 3 x^{2} \Rightarrow 3 x^{2} + 11 x - 4 = 0 \Rightarrow 3 x^{2} + 12 x - x - 4 = 0 \Rightarrow 3 x (x + 4) - 1 (x + 4) = 0 \Rightarrow (x + 4) (3 x - 1) = 0 \Rightarrow x + 4 = 0 or 3 x - 1 = 0 \Rightarrow x = - 4 or x = \frac{1}{3} H ence, the roots of the equation are - 4 and \frac{1}{3} .$

Answer:

$Given: 48 x^{2} - 13 x - 1 = 0 \Rightarrow 48 x^{2} - (16 x - 3 x) - 1 = 0 \Rightarrow 48 x^{2} - 16 x + 3 x - 1 = 0 \Rightarrow 16 x (3 x - 1) + 1 (3 x - 1) = 0 \Rightarrow (16 x + 1) (3 x - 1) = 0 \Rightarrow 16 x + 1 = 0 or 3 x - 1 = 0 \Rightarrow x = \frac{- 1}{16} or x = \frac{1}{3} Hence, the roots of the equation are \frac{- 1}{16} and \frac{1}{3} .$

Question 20:

$Given: 48 x^{2} - 13 x - 1 = 0 \Rightarrow 48 x^{2} - (16 x - 3 x) - 1 = 0 \Rightarrow 48 x^{2} - 16 x + 3 x - 1 = 0 \Rightarrow 16 x (3 x - 1) + 1 (3 x - 1) = 0 \Rightarrow (16 x + 1) (3 x - 1) = 0 \Rightarrow 16 x + 1 = 0 or 3 x - 1 = 0 \Rightarrow x = \frac{- 1}{16} or x = \frac{1}{3} Hence, the roots of the equation are \frac{- 1}{16} and \frac{1}{3} .$

Answer:

We write, $2 \sqrt{2} x = 3 \sqrt{2} x - \sqrt{2} x$ as $x^{2} \times (- 6) = - 6 x^{2} = 3 \sqrt{2} x \times (- \sqrt{2} x)$

$∴ x^{2} + 2 \sqrt{2} x - 6 = 0 \Rightarrow x^{2} + 3 \sqrt{2} x - \sqrt{2} x - 6 = 0 \Rightarrow x (x + 3 \sqrt{2}) - \sqrt{2} (x + 3 \sqrt{2}) = 0 \Rightarrow (x + 3 \sqrt{2}) (x - \sqrt{2}) = 0$
$\Rightarrow x + 3 \sqrt{2} = 0 or x - \sqrt{2} = 0 \Rightarrow x = - 3 \sqrt{2} or x = \sqrt{2}$
Hence, the roots of the given equation are $- 3 \sqrt{2}$ and $\sqrt{2}$ .

Question 21:

We write, $2 \sqrt{2} x = 3 \sqrt{2} x - \sqrt{2} x$ as $x^{2} \times (- 6) = - 6 x^{2} = 3 \sqrt{2} x \times (- \sqrt{2} x)$

$∴ x^{2} + 2 \sqrt{2} x - 6 = 0 \Rightarrow x^{2} + 3 \sqrt{2} x - \sqrt{2} x - 6 = 0 \Rightarrow x (x + 3 \sqrt{2}) - \sqrt{2} (x + 3 \sqrt{2}) = 0 \Rightarrow (x + 3 \sqrt{2}) (x - \sqrt{2}) = 0$
$\Rightarrow x + 3 \sqrt{2} = 0 or x - \sqrt{2} = 0 \Rightarrow x = - 3 \sqrt{2} or x = \sqrt{2}$
Hence, the roots of the given equation are $- 3 \sqrt{2}$ and $\sqrt{2}$ .

Answer:

Consider $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$
Factorising by splitting the middle term;
$\sqrt{3} x^{2} + 12 x - 2 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 4 \sqrt{3}) - 2 (x + 4 \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x - 2) (x + 4 \sqrt{3}) = 0 \Rightarrow \sqrt{3} x - 2 = 0 or x + 4 \sqrt{3} = 0 \Rightarrow x = \frac{2}{\sqrt{3}} or x = - 4 \sqrt{3}$
Hence, the roots of the given equation are $\frac{2}{\sqrt{3}}$ and $- 4 \sqrt{3}$ .

Question 22:

Consider $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$
Factorising by splitting the middle term;
$\sqrt{3} x^{2} + 12 x - 2 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 4 \sqrt{3}) - 2 (x + 4 \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x - 2) (x + 4 \sqrt{3}) = 0 \Rightarrow \sqrt{3} x - 2 = 0 or x + 4 \sqrt{3} = 0 \Rightarrow x = \frac{2}{\sqrt{3}} or x = - 4 \sqrt{3}$
Hence, the roots of the given equation are $\frac{2}{\sqrt{3}}$ and $- 4 \sqrt{3}$ .

Answer:

$Given : \sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0 \Rightarrow (x + 3 \sqrt{3}) (\sqrt{3} x + 2) = 0 \Rightarrow x + 3 \sqrt{3} = 0 or \sqrt{3} x + 2 = 0 \Rightarrow x = - 3 \sqrt{3} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{- 2 \sqrt{3}}{3} Hence, the roots of the equation are - 3 \sqrt{3} and \frac{- 2 \sqrt{3}}{3} .$

Question 23:

$Given : \sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0 \Rightarrow (x + 3 \sqrt{3}) (\sqrt{3} x + 2) = 0 \Rightarrow x + 3 \sqrt{3} = 0 or \sqrt{3} x + 2 = 0 \Rightarrow x = - 3 \sqrt{3} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{- 2 \sqrt{3}}{3} Hence, the roots of the equation are - 3 \sqrt{3} and \frac{- 2 \sqrt{3}}{3} .$

Answer:

$Given: 3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0 \Rightarrow 3 \sqrt{7} x^{2} + 7 x - 3 x - \sqrt{7} = 0 \Rightarrow \sqrt{7} x (3 x + \sqrt{7}) - 1 (3 x + \sqrt{7}) = 0 \Rightarrow (3 x + \sqrt{7}) (\sqrt{7} x - 1) = 0 \Rightarrow 3 x + \sqrt{7} = 0 or \sqrt{7} x - 1 = 0 \Rightarrow x = \frac{- \sqrt{7}}{3} or x = \frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7} H ence, the roots of the equation are \frac{- \sqrt{7}}{3} and \frac{\sqrt{7}}{7} .$

Question 24:

$Given: 3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0 \Rightarrow 3 \sqrt{7} x^{2} + 7 x - 3 x - \sqrt{7} = 0 \Rightarrow \sqrt{7} x (3 x + \sqrt{7}) - 1 (3 x + \sqrt{7}) = 0 \Rightarrow (3 x + \sqrt{7}) (\sqrt{7} x - 1) = 0 \Rightarrow 3 x + \sqrt{7} = 0 or \sqrt{7} x - 1 = 0 \Rightarrow x = \frac{- \sqrt{7}}{3} or x = \frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7} H ence, the roots of the equation are \frac{- \sqrt{7}}{3} and \frac{\sqrt{7}}{7} .$

Answer:

We write, $- 6 x = 7 x - 13 x$ as $\sqrt{7} x^{2} \times (- 13 \sqrt{7}) = - 91 x^{2} = 7 x \times (- 13 x)$

$∴ \sqrt{7} x^{2} - 6 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x^{2} + 7 x - 13 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x (x + \sqrt{7}) - 13 (x + \sqrt{7}) = 0 \Rightarrow (x + \sqrt{7}) (\sqrt{7} x - 13) = 0$
$\Rightarrow x + \sqrt{7} = 0 or \sqrt{7} x - 13 = 0 \Rightarrow x = - \sqrt{7} or x = \frac{13}{\sqrt{7}} = \frac{13 \sqrt{7}}{7}$
Hence, the roots of the given equation are $- \sqrt{7}$ and $\frac{13 \sqrt{7}}{7}$ .

Question 25:

We write, $- 6 x = 7 x - 13 x$ as $\sqrt{7} x^{2} \times (- 13 \sqrt{7}) = - 91 x^{2} = 7 x \times (- 13 x)$

$∴ \sqrt{7} x^{2} - 6 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x^{2} + 7 x - 13 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x (x + \sqrt{7}) - 13 (x + \sqrt{7}) = 0 \Rightarrow (x + \sqrt{7}) (\sqrt{7} x - 13) = 0$
$\Rightarrow x + \sqrt{7} = 0 or \sqrt{7} x - 13 = 0 \Rightarrow x = - \sqrt{7} or x = \frac{13}{\sqrt{7}} = \frac{13 \sqrt{7}}{7}$
Hence, the roots of the given equation are $- \sqrt{7}$ and $\frac{13 \sqrt{7}}{7}$ .

Answer:

$Given: 4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{6} x^{2} - 16 x + 3 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{2} x (\sqrt{3} x - 2 \sqrt{2}) + \sqrt{3} (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow (4 \sqrt{2} x + \sqrt{3}) (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow 4 \sqrt{2} x + \sqrt{3} = 0 or \sqrt{3} x - 2 \sqrt{2} = 0 \Rightarrow x = \frac{- \sqrt{3}}{4 \sqrt{2}} = \frac{- \sqrt{3} \times \sqrt{2}}{4 \sqrt{2} \times \sqrt{2}} = \frac{- \sqrt{6}}{8} or x = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2 \sqrt{6}}{3} H ence, the roots of the equation are \frac{- \sqrt{6}}{8} and \frac{2 \sqrt{6}}{3} .$

Question 26:

$Given: 4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{6} x^{2} - 16 x + 3 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{2} x (\sqrt{3} x - 2 \sqrt{2}) + \sqrt{3} (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow (4 \sqrt{2} x + \sqrt{3}) (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow 4 \sqrt{2} x + \sqrt{3} = 0 or \sqrt{3} x - 2 \sqrt{2} = 0 \Rightarrow x = \frac{- \sqrt{3}}{4 \sqrt{2}} = \frac{- \sqrt{3} \times \sqrt{2}}{4 \sqrt{2} \times \sqrt{2}} = \frac{- \sqrt{6}}{8} or x = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2 \sqrt{6}}{3} H ence, the roots of the equation are \frac{- \sqrt{6}}{8} and \frac{2 \sqrt{6}}{3} .$

Answer:

We write, $- 2 \sqrt{6} x = - \sqrt{6} x - \sqrt{6} x$ as $3 x^{2} \times 2 = 6 x^{2} = (- \sqrt{6} x) \times (- \sqrt{6} x)$

$∴ 3 x^{2} - 2 \sqrt{6} x + 2 = 0 \Rightarrow 3 x^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \Rightarrow \sqrt{3} x (\sqrt{3} x - \sqrt{2}) - \sqrt{2} (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow (\sqrt{3} x - \sqrt{2}) (\sqrt{3} x - \sqrt{2}) = 0$
$\Rightarrow {(\sqrt{3} x - \sqrt{2})}^{2} = 0 \Rightarrow \sqrt{3} x - \sqrt{2} = 0 \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$
Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

Question 27:

We write, $- 2 \sqrt{6} x = - \sqrt{6} x - \sqrt{6} x$ as $3 x^{2} \times 2 = 6 x^{2} = (- \sqrt{6} x) \times (- \sqrt{6} x)$

$∴ 3 x^{2} - 2 \sqrt{6} x + 2 = 0 \Rightarrow 3 x^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \Rightarrow \sqrt{3} x (\sqrt{3} x - \sqrt{2}) - \sqrt{2} (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow (\sqrt{3} x - \sqrt{2}) (\sqrt{3} x - \sqrt{2}) = 0$
$\Rightarrow {(\sqrt{3} x - \sqrt{2})}^{2} = 0 \Rightarrow \sqrt{3} x - \sqrt{2} = 0 \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$
Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

Answer:

We write, $- 2 \sqrt{2} x = - 3 \sqrt{2} x + \sqrt{2} x$ as $\sqrt{3} x^{2} \times (- 2 \sqrt{3}) = - 6 x^{2} = (- 3 \sqrt{2} x) \times (\sqrt{2} x)$

$∴ \sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (x - \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$
$\Rightarrow x - \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0 \Rightarrow x = \sqrt{6} or x = - \frac{\sqrt{2}}{\sqrt{3}} = - \frac{\sqrt{6}}{3}$
Hence, the roots of the given equation are $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ .

Question 28:

We write, $- 2 \sqrt{2} x = - 3 \sqrt{2} x + \sqrt{2} x$ as $\sqrt{3} x^{2} \times (- 2 \sqrt{3}) = - 6 x^{2} = (- 3 \sqrt{2} x) \times (\sqrt{2} x)$

$∴ \sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (x - \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$
$\Rightarrow x - \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0 \Rightarrow x = \sqrt{6} or x = - \frac{\sqrt{2}}{\sqrt{3}} = - \frac{\sqrt{6}}{3}$
Hence, the roots of the given equation are $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ .

Answer:

We write, $- 3 \sqrt{5} x = - 2 \sqrt{5} x - \sqrt{5} x$ as $x^{2} \times 10 = 10 x^{2} = (- 2 \sqrt{5} x) \times (- \sqrt{5} x)$

$∴ x^{2} - 3 \sqrt{5} x + 10 = 0 \Rightarrow x^{2} - 2 \sqrt{5} x - \sqrt{5} x + 10 = 0 \Rightarrow x (x - 2 \sqrt{5}) - \sqrt{5} (x - 2 \sqrt{5}) = 0 \Rightarrow (x - 2 \sqrt{5}) (x - \sqrt{5}) = 0$
$\Rightarrow x - \sqrt{5} = 0 or x - 2 \sqrt{5} = 0 \Rightarrow x = \sqrt{5} or x = 2 \sqrt{5}$
Hence, the roots of the given equation are $\sqrt{5}$ and $2 \sqrt{5}$ .

Question 29:

We write, $- 3 \sqrt{5} x = - 2 \sqrt{5} x - \sqrt{5} x$ as $x^{2} \times 10 = 10 x^{2} = (- 2 \sqrt{5} x) \times (- \sqrt{5} x)$

$∴ x^{2} - 3 \sqrt{5} x + 10 = 0 \Rightarrow x^{2} - 2 \sqrt{5} x - \sqrt{5} x + 10 = 0 \Rightarrow x (x - 2 \sqrt{5}) - \sqrt{5} (x - 2 \sqrt{5}) = 0 \Rightarrow (x - 2 \sqrt{5}) (x - \sqrt{5}) = 0$
$\Rightarrow x - \sqrt{5} = 0 or x - 2 \sqrt{5} = 0 \Rightarrow x = \sqrt{5} or x = 2 \sqrt{5}$
Hence, the roots of the given equation are $\sqrt{5}$ and $2 \sqrt{5}$ .

Answer:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0 \Rightarrow x^{2} - \sqrt{3} x - x + \sqrt{3} = 0 \Rightarrow x (x - \sqrt{3}) - 1 (x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3}) (x - 1) = 0$
$\Rightarrow x - \sqrt{3} = 0 or x - 1 = 0 \Rightarrow x = \sqrt{3} or x = 1$
Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Question 30:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0 \Rightarrow x^{2} - \sqrt{3} x - x + \sqrt{3} = 0 \Rightarrow x (x - \sqrt{3}) - 1 (x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3}) (x - 1) = 0$
$\Rightarrow x - \sqrt{3} = 0 or x - 1 = 0 \Rightarrow x = \sqrt{3} or x = 1$
Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Answer:

We write, $3 \sqrt{3} x = 5 \sqrt{3} x - 2 \sqrt{3} x$ as $x^{2} \times (- 30) = - 30 x^{2} = 5 \sqrt{3} x \times (- 2 \sqrt{3} x)$

$∴ x^{2} + 3 \sqrt{3} x - 30 = 0 \Rightarrow x^{2} + 5 \sqrt{3} x - 2 \sqrt{3} x - 30 = 0 \Rightarrow x (x + 5 \sqrt{3}) - 2 \sqrt{3} (x + 5 \sqrt{3}) = 0 \Rightarrow (x + 5 \sqrt{3}) (x - 2 \sqrt{3}) = 0$
$\Rightarrow x + 5 \sqrt{3} = 0 or x - 2 \sqrt{3} = 0 \Rightarrow x = - 5 \sqrt{3} or x = 2 \sqrt{3}$
Hence, the roots of the given equation are $- 5 \sqrt{3}$ and $2 \sqrt{3}$ .

Question 31:

We write, $3 \sqrt{3} x = 5 \sqrt{3} x - 2 \sqrt{3} x$ as $x^{2} \times (- 30) = - 30 x^{2} = 5 \sqrt{3} x \times (- 2 \sqrt{3} x)$

$∴ x^{2} + 3 \sqrt{3} x - 30 = 0 \Rightarrow x^{2} + 5 \sqrt{3} x - 2 \sqrt{3} x - 30 = 0 \Rightarrow x (x + 5 \sqrt{3}) - 2 \sqrt{3} (x + 5 \sqrt{3}) = 0 \Rightarrow (x + 5 \sqrt{3}) (x - 2 \sqrt{3}) = 0$
$\Rightarrow x + 5 \sqrt{3} = 0 or x - 2 \sqrt{3} = 0 \Rightarrow x = - 5 \sqrt{3} or x = 2 \sqrt{3}$
Hence, the roots of the given equation are $- 5 \sqrt{3}$ and $2 \sqrt{3}$ .

Answer:

We write, $7 x = 5 x + 2 x$ as $\sqrt{2} x^{2} \times 5 \sqrt{2} = 10 x^{2} = 5 x \times 2 x$

$∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 \Rightarrow x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \Rightarrow (\sqrt{2} x + 5) (x + \sqrt{2}) = 0$
$\Rightarrow x + \sqrt{2} = 0 or \sqrt{2} x + 5 = 0 \Rightarrow x = - \sqrt{2} or x = - \frac{5}{\sqrt{2}} = - \frac{5 \sqrt{2}}{2}$
Hence, the roots of the given equation are $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ .

Question 32:

We write, $7 x = 5 x + 2 x$ as $\sqrt{2} x^{2} \times 5 \sqrt{2} = 10 x^{2} = 5 x \times 2 x$

$∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 \Rightarrow x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \Rightarrow (\sqrt{2} x + 5) (x + \sqrt{2}) = 0$
$\Rightarrow x + \sqrt{2} = 0 or \sqrt{2} x + 5 = 0 \Rightarrow x = - \sqrt{2} or x = - \frac{5}{\sqrt{2}} = - \frac{5 \sqrt{2}}{2}$
Hence, the roots of the given equation are $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ .

Answer:

We write, 13x = 5x + 8x as $5 x^{2} \times 8 = 40 x^{2} = 5 x \times 8 x$
$∴ 5 x^{2} + 13 x + 8 = 0 \Rightarrow 5 x^{2} + 5 x + 8 x + 8 = 0 \Rightarrow 5 x (x + 1) + 8 (x + 1) = 0 \Rightarrow (x + 1) (5 x + 8) = 0$
$\Rightarrow x + 1 = 0 or 5 x + 8 = 0 \Rightarrow x = - 1 or x = - \frac{8}{5}$
Hence, $- 1$ and $- \frac{8}{5}$ are the roots of the given equation.

Question 33:

We write, 13x = 5x + 8x as $5 x^{2} \times 8 = 40 x^{2} = 5 x \times 8 x$
$∴ 5 x^{2} + 13 x + 8 = 0 \Rightarrow 5 x^{2} + 5 x + 8 x + 8 = 0 \Rightarrow 5 x (x + 1) + 8 (x + 1) = 0 \Rightarrow (x + 1) (5 x + 8) = 0$
$\Rightarrow x + 1 = 0 or 5 x + 8 = 0 \Rightarrow x = - 1 or x = - \frac{8}{5}$
Hence, $- 1$ and $- \frac{8}{5}$ are the roots of the given equation.

Answer:

$x^{2} - (1 + \sqrt{2}) x + \sqrt{2} = 0 x^{2} - x - \sqrt{2} x + \sqrt{2} = 0 x (x - 1) - \sqrt{2} (x - 1) = 0 (x - \sqrt{2}) (x - 1) = 0 \Rightarrow x - \sqrt{2} = 0 and x - 1 = 0 \Rightarrow x = \sqrt{2} and x = 1$

Question 34:

$x^{2} - (1 + \sqrt{2}) x + \sqrt{2} = 0 x^{2} - x - \sqrt{2} x + \sqrt{2} = 0 x (x - 1) - \sqrt{2} (x - 1) = 0 (x - \sqrt{2}) (x - 1) = 0 \Rightarrow x - \sqrt{2} = 0 and x - 1 = 0 \Rightarrow x = \sqrt{2} and x = 1$

Answer:

$Given : 9 x^{2} + 6 x + 1 = 0 \Rightarrow 9 x^{2} + 3 x + 3 x + 1 = 0 \Rightarrow 3 x (3 x + 1) + 1 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x + 1) = 0 \Rightarrow 3 x + 1 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 1}{3} Hence, \frac{- 1}{3} is the root of the equation 9 x^{2} + 6 x + 1 = 0 .$

Question 35:

$Given : 9 x^{2} + 6 x + 1 = 0 \Rightarrow 9 x^{2} + 3 x + 3 x + 1 = 0 \Rightarrow 3 x (3 x + 1) + 1 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x + 1) = 0 \Rightarrow 3 x + 1 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 1}{3} Hence, \frac{- 1}{3} is the root of the equation 9 x^{2} + 6 x + 1 = 0 .$

Answer:

We write, $- 20 x = - 10 x - 10 x$ as $100 x^{2} \times 1 = 100 x^{2} = (- 10 x) \times (- 10 x)$
$∴ 100 x^{2} - 20 x + 1 = 0 \Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0 \Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0 \Rightarrow (10 x - 1) (10 x - 1) = 0$
$\Rightarrow {(10 x - 1)}^{2} = 0 \Rightarrow 10 x - 1 = 0 \Rightarrow x = \frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

Question 36:

We write, $- 20 x = - 10 x - 10 x$ as $100 x^{2} \times 1 = 100 x^{2} = (- 10 x) \times (- 10 x)$
$∴ 100 x^{2} - 20 x + 1 = 0 \Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0 \Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0 \Rightarrow (10 x - 1) (10 x - 1) = 0$
$\Rightarrow {(10 x - 1)}^{2} = 0 \Rightarrow 10 x - 1 = 0 \Rightarrow x = \frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

Answer:

We write, $- x = - \frac{x}{2} - \frac{x}{2}$ as $2 x^{2} \times \frac{1}{8} = \frac{x^{2}}{4} = (- \frac{x}{2}) \times (- \frac{x}{2})$
$∴ 2 x^{2} - x + \frac{1}{8} = 0 \Rightarrow 2 x^{2} - \frac{x}{2} - \frac{x}{2} + \frac{1}{8} = 0 \Rightarrow 2 x (x - \frac{1}{4}) - \frac{1}{2} (x - \frac{1}{4}) = 0 \Rightarrow (x - \frac{1}{4}) (2 x - \frac{1}{2}) = 0$
$\Rightarrow x - \frac{1}{4} = 0 or 2 x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{4} or x = \frac{1}{4}$
Hence, $\frac{1}{4}$ is the repeated root of the given equation.

Question 37:

We write, $- x = - \frac{x}{2} - \frac{x}{2}$ as $2 x^{2} \times \frac{1}{8} = \frac{x^{2}}{4} = (- \frac{x}{2}) \times (- \frac{x}{2})$
$∴ 2 x^{2} - x + \frac{1}{8} = 0 \Rightarrow 2 x^{2} - \frac{x}{2} - \frac{x}{2} + \frac{1}{8} = 0 \Rightarrow 2 x (x - \frac{1}{4}) - \frac{1}{2} (x - \frac{1}{4}) = 0 \Rightarrow (x - \frac{1}{4}) (2 x - \frac{1}{2}) = 0$
$\Rightarrow x - \frac{1}{4} = 0 or 2 x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{4} or x = \frac{1}{4}$
Hence, $\frac{1}{4}$ is the repeated root of the given equation.

Answer:

$Given: 10 x - \frac{1}{x} = 3 \Rightarrow 10 x^{2} - 1 = 3 x [Multiplying both sides by x] \Rightarrow 10 x^{2} - 3 x - 1 = 0 \Rightarrow 10 x^{2} - (5 x - 2 x) - 1 = 0 \Rightarrow 10 x^{2} - 5 x + 2 x - 1 = 0 \Rightarrow 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0 \Rightarrow 2 x - 1 = 0 or 5 x + 1 = 0 \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{5} H ence, the roots of the equation are \frac{1}{2} and \frac{- 1}{5} .$

Question 38:

$Given: 10 x - \frac{1}{x} = 3 \Rightarrow 10 x^{2} - 1 = 3 x [Multiplying both sides by x] \Rightarrow 10 x^{2} - 3 x - 1 = 0 \Rightarrow 10 x^{2} - (5 x - 2 x) - 1 = 0 \Rightarrow 10 x^{2} - 5 x + 2 x - 1 = 0 \Rightarrow 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0 \Rightarrow 2 x - 1 = 0 or 5 x + 1 = 0 \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{5} H ence, the roots of the equation are \frac{1}{2} and \frac{- 1}{5} .$

Answer:

$Given: \frac{2}{x^{2}} - \frac{5}{x} + 2 = 0 \Rightarrow 2 - 5 x + 2 x^{2} = 0 [Multiplying both side by x^{2}] \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow 2 x^{2} - (4 x + x) + 2 = 0 \Rightarrow 2 x^{2} - 4 x - x + 2 = 0 \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \Rightarrow (2 x - 1) (x - 2) = 0 \Rightarrow 2 x - 1 = 0 or x - 2 = 0 \Rightarrow x = \frac{1}{2} or x = 2 H ence, the roots of the equation are \frac{1}{2} and 2 .$

Question 39:

$Given: \frac{2}{x^{2}} - \frac{5}{x} + 2 = 0 \Rightarrow 2 - 5 x + 2 x^{2} = 0 [Multiplying both side by x^{2}] \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow 2 x^{2} - (4 x + x) + 2 = 0 \Rightarrow 2 x^{2} - 4 x - x + 2 = 0 \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \Rightarrow (2 x - 1) (x - 2) = 0 \Rightarrow 2 x - 1 = 0 or x - 2 = 0 \Rightarrow x = \frac{1}{2} or x = 2 H ence, the roots of the equation are \frac{1}{2} and 2 .$

Answer:

We write, $a x = 2 a x - a x$ as $2 x^{2} \times (- a^{2}) = - 2 a^{2} x^{2} = 2 a x \times (- a x)$
$∴ 2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (x + a) (2 x - a) = 0$
$\Rightarrow x + a = 0 or 2 x - a = 0 \Rightarrow x = - a or x = \frac{a}{2}$
Hence, $- a$ and $\frac{a}{2}$ are the roots of the given equation.

Question 40:

We write, $a x = 2 a x - a x$ as $2 x^{2} \times (- a^{2}) = - 2 a^{2} x^{2} = 2 a x \times (- a x)$
$∴ 2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (x + a) (2 x - a) = 0$
$\Rightarrow x + a = 0 or 2 x - a = 0 \Rightarrow x = - a or x = \frac{a}{2}$
Hence, $- a$ and $\frac{a}{2}$ are the roots of the given equation.

Answer:

We write, $4 b x = 2 (a + b) x - 2 (a - b) x$ as $4 x^{2} \times [- (a^{2} - b^{2})] = - 4 (a^{2} - b^{2}) x^{2} = 2 (a + b) x \times [- 2 (a - b) x]$
$∴ 4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0 \Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0$
$\Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$
Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

Question 41:

We write, $4 b x = 2 (a + b) x - 2 (a - b) x$ as $4 x^{2} \times [- (a^{2} - b^{2})] = - 4 (a^{2} - b^{2}) x^{2} = 2 (a + b) x \times [- 2 (a - b) x]$
$∴ 4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0 \Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0$
$\Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$
Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

Answer:

We write, $- 4 a^{2} x = - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x$ as $4 x^{2} \times (a^{4} - b^{4}) = 4 (a^{4} - b^{4}) x^{2} = [- 2 (a^{2} + b^{2})] x \times [- 2 (a^{2} - b^{2})] x$
$∴ 4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0 \Rightarrow 4 x^{2} - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x + (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 2 x [2 x - (a^{2} + b^{2})] - (a^{2} - b^{2}) [2 x - (a^{2} + b^{2})] = 0 \Rightarrow [2 x - (a^{2} + b^{2})] [2 x - (a^{2} - b^{2})] = 0$
$\Rightarrow 2 x - (a^{2} + b^{2}) = 0 or 2 x - (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{2} or x = \frac{a^{2} - b^{2}}{2}$
Hence, $\frac{a^{2} + b^{2}}{2}$ and $\frac{a^{2} - b^{2}}{2}$ are the roots of the given equation.

Question 42:

We write, $- 4 a^{2} x = - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x$ as $4 x^{2} \times (a^{4} - b^{4}) = 4 (a^{4} - b^{4}) x^{2} = [- 2 (a^{2} + b^{2})] x \times [- 2 (a^{2} - b^{2})] x$
$∴ 4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0 \Rightarrow 4 x^{2} - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x + (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 2 x [2 x - (a^{2} + b^{2})] - (a^{2} - b^{2}) [2 x - (a^{2} + b^{2})] = 0 \Rightarrow [2 x - (a^{2} + b^{2})] [2 x - (a^{2} - b^{2})] = 0$
$\Rightarrow 2 x - (a^{2} + b^{2}) = 0 or 2 x - (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{2} or x = \frac{a^{2} - b^{2}}{2}$
Hence, $\frac{a^{2} + b^{2}}{2}$ and $\frac{a^{2} - b^{2}}{2}$ are the roots of the given equation.

Answer:

We write, $5 x = (a + 3) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + a - 6)] = - (a^{2} + a - 6) x^{2} = (a + 3) x \times [- (a - 2) x]$
$∴ x^{2} + 5 x - (a^{2} + a - 6) = 0 \Rightarrow x^{2} + (a + 3) x - (a - 2) x - (a + 3) (a - 2) = 0 \Rightarrow x [x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 3) or x = a - 2$
Hence, $- (a + 3)$ and $(a - 2)$ are the roots of the given equation.

Question 43:

We write, $5 x = (a + 3) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + a - 6)] = - (a^{2} + a - 6) x^{2} = (a + 3) x \times [- (a - 2) x]$
$∴ x^{2} + 5 x - (a^{2} + a - 6) = 0 \Rightarrow x^{2} + (a + 3) x - (a - 2) x - (a + 3) (a - 2) = 0 \Rightarrow x [x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 3) or x = a - 2$
Hence, $- (a + 3)$ and $(a - 2)$ are the roots of the given equation.

Answer:

We write, $- 2 a x = (2 b - a) x - (2 b + a) x$ as $x^{2} \times [- (4 b^{2} - a^{2})] = - (4 b^{2} - a^{2}) x^{2} = (2 b - a) x \times [- (2 b + a) x]$
$∴ x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0 \Rightarrow x^{2} + (2 b - a) x - (2 b + a) x - (2 b - a) (2 b + a) = 0 \Rightarrow x [x + (2 b - a)] - (2 b + a) [x + (2 b - a)] = 0 \Rightarrow [x + (2 b - a)] [x - (2 b + a)] = 0$
$\Rightarrow x + (2 b - a) = 0 or x - (2 b + a) = 0 \Rightarrow x = - (2 b - a) or x = 2 b + a \Rightarrow x = a - 2 b or x = a + 2 b$
Hence, $a - 2 b$ and $a + 2 b$ are the roots of the given equation.

Question 44:

We write, $- 2 a x = (2 b - a) x - (2 b + a) x$ as $x^{2} \times [- (4 b^{2} - a^{2})] = - (4 b^{2} - a^{2}) x^{2} = (2 b - a) x \times [- (2 b + a) x]$
$∴ x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0 \Rightarrow x^{2} + (2 b - a) x - (2 b + a) x - (2 b - a) (2 b + a) = 0 \Rightarrow x [x + (2 b - a)] - (2 b + a) [x + (2 b - a)] = 0 \Rightarrow [x + (2 b - a)] [x - (2 b + a)] = 0$
$\Rightarrow x + (2 b - a) = 0 or x - (2 b + a) = 0 \Rightarrow x = - (2 b - a) or x = 2 b + a \Rightarrow x = a - 2 b or x = a + 2 b$
Hence, $a - 2 b$ and $a + 2 b$ are the roots of the given equation.

Answer:

We write, $- (2 b - 1) x = - (b - 5) x - (b + 4) x$ as $x^{2} \times (b^{2} - b - 20) = (b^{2} - b - 20) x^{2} = [- (b - 5) x] \times [- (b + 4) x]$
$∴ x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0 \Rightarrow x^{2} - (b - 5) x - (b + 4) x + (b - 5) (b + 4) = 0 \Rightarrow x [x - (b - 5)] - (b + 4) [x - (b - 5)] = 0 \Rightarrow [x - (b - 5)] [x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0 \Rightarrow x = b - 5 or x = b + 4$
Hence, $b - 5$ and $b + 4$ are the roots of the given equation.

Question 45:

We write, $- (2 b - 1) x = - (b - 5) x - (b + 4) x$ as $x^{2} \times (b^{2} - b - 20) = (b^{2} - b - 20) x^{2} = [- (b - 5) x] \times [- (b + 4) x]$
$∴ x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0 \Rightarrow x^{2} - (b - 5) x - (b + 4) x + (b - 5) (b + 4) = 0 \Rightarrow x [x - (b - 5)] - (b + 4) [x - (b - 5)] = 0 \Rightarrow [x - (b - 5)] [x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0 \Rightarrow x = b - 5 or x = b + 4$
Hence, $b - 5$ and $b + 4$ are the roots of the given equation.

Answer:

We write, $6 x = (a + 4) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + 2 a - 8)] = - (a^{2} + 2 a - 8) x^{2} = (a + 4) x \times [- (a - 2) x]$
$∴ x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0 \Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = a - 2$
Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

Question 46:

We write, $6 x = (a + 4) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + 2 a - 8)] = - (a^{2} + 2 a - 8) x^{2} = (a + 4) x \times [- (a - 2) x]$
$∴ x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0 \Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = a - 2$
Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

Answer:

$Given: a b x^{2} + (b^{2} - a c) x - b c = 0 \Rightarrow a b x^{2} + b^{2} x - a c x - b c = 0 \Rightarrow b x (a x + b) - c (a x + b) = 0 \Rightarrow (b x - c) (a x + b) = 0 \Rightarrow b x - c = 0 or a x + b = 0 \Rightarrow x = \frac{c}{b} or x = \frac{- b}{a} H ence, the roots of the equation are \frac{c}{b} and \frac{- b}{a} .$

Question 47:

$Given: a b x^{2} + (b^{2} - a c) x - b c = 0 \Rightarrow a b x^{2} + b^{2} x - a c x - b c = 0 \Rightarrow b x (a x + b) - c (a x + b) = 0 \Rightarrow (b x - c) (a x + b) = 0 \Rightarrow b x - c = 0 or a x + b = 0 \Rightarrow x = \frac{c}{b} or x = \frac{- b}{a} H ence, the roots of the equation are \frac{c}{b} and \frac{- b}{a} .$

Answer:

We write, $- 4 a x = - (b + 2 a) x + (b - 2 a) x$ as $x^{2} \times (- b^{2} + 4 a^{2}) = (- b^{2} + 4 a^{2}) x^{2} = - (b + 2 a) x \times (b - 2 a) x$
$∴ x^{2} - 4 a x - b^{2} + 4 a^{2} = 0 \Rightarrow x^{2} - (b + 2 a) x + (b - 2 a) x - (b - 2 a) (b + 2 a) = 0 \Rightarrow x [x - (b + 2 a)] + (b - 2 a) [x - (b + 2 a)] = 0 \Rightarrow [x - (b + 2 a)] [x + (b - 2 a)] = 0$
$\Rightarrow x - (b + 2 a) = 0 or x + (b - 2 a) = 0 \Rightarrow x = 2 a + b or x = - (b - 2 a) \Rightarrow x = 2 a + b or x = 2 a - b$
Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Question 48:

We write, $- 4 a x = - (b + 2 a) x + (b - 2 a) x$ as $x^{2} \times (- b^{2} + 4 a^{2}) = (- b^{2} + 4 a^{2}) x^{2} = - (b + 2 a) x \times (b - 2 a) x$
$∴ x^{2} - 4 a x - b^{2} + 4 a^{2} = 0 \Rightarrow x^{2} - (b + 2 a) x + (b - 2 a) x - (b - 2 a) (b + 2 a) = 0 \Rightarrow x [x - (b + 2 a)] + (b - 2 a) [x - (b + 2 a)] = 0 \Rightarrow [x - (b + 2 a)] [x + (b - 2 a)] = 0$
$\Rightarrow x - (b + 2 a) = 0 or x + (b - 2 a) = 0 \Rightarrow x = 2 a + b or x = - (b - 2 a) \Rightarrow x = 2 a + b or x = 2 a - b$
Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Answer:

$Given : 4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0 \Rightarrow 4 x^{2} - 2 a^{2} x - 2 b^{2} x + a^{2} b^{2} = 0 \Rightarrow 2 x (2 x - a^{2}) - b^{2} (2 x - a^{2}) = 0 \Rightarrow (2 x - b^{2}) (2 x - a^{2}) = 0 \Rightarrow 2 x - b^{2} = 0 or 2 x - a^{2} = 0 \Rightarrow x = \frac{b^{2}}{2} or x = \frac{a^{2}}{2} H ence, the roots of the equation are \frac{b^{2}}{2} and \frac{a^{2}}{2} .$

Question 49:

$Given : 4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0 \Rightarrow 4 x^{2} - 2 a^{2} x - 2 b^{2} x + a^{2} b^{2} = 0 \Rightarrow 2 x (2 x - a^{2}) - b^{2} (2 x - a^{2}) = 0 \Rightarrow (2 x - b^{2}) (2 x - a^{2}) = 0 \Rightarrow 2 x - b^{2} = 0 or 2 x - a^{2} = 0 \Rightarrow x = \frac{b^{2}}{2} or x = \frac{a^{2}}{2} H ence, the roots of the equation are \frac{b^{2}}{2} and \frac{a^{2}}{2} .$

Answer:

$Given : 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 \Rightarrow 12 a b x^{2} - 9 a^{2} x + 8 b^{2} x - 6 a b = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (3 a x + 2 b) (4 b x - 3 a) = 0 \Rightarrow 3 a x + 2 b = 0 or 4 b x - 3 a = 0 \Rightarrow x = \frac{- 2 b}{3 a} or x = \frac{3 a}{4 b} H ence, the roots of the equation are \frac{- 2 b}{3 a} and \frac{3 a}{4 b} .$

Question 50:

$Given : 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 \Rightarrow 12 a b x^{2} - 9 a^{2} x + 8 b^{2} x - 6 a b = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (3 a x + 2 b) (4 b x - 3 a) = 0 \Rightarrow 3 a x + 2 b = 0 or 4 b x - 3 a = 0 \Rightarrow x = \frac{- 2 b}{3 a} or x = \frac{3 a}{4 b} H ence, the roots of the equation are \frac{- 2 b}{3 a} and \frac{3 a}{4 b} .$

Answer:

$\begin{array}{l} Given : \\ a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0 \Rightarrow b^{2} x (a^{2} x + 1) - 1 (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) = 0 or (a^{2} x + 1) = 0 \Rightarrow x = \frac{1}{b^{2}} or x = \frac{- 1}{a^{2}} H ence, \frac{1}{b^{2}} and \frac{- 1}{a^{2}} are the roots of the given equation. \end{array}$

Question 51:

$\begin{array}{l} Given : \\ a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0 \Rightarrow b^{2} x (a^{2} x + 1) - 1 (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) = 0 or (a^{2} x + 1) = 0 \Rightarrow x = \frac{1}{b^{2}} or x = \frac{- 1}{a^{2}} H ence, \frac{1}{b^{2}} and \frac{- 1}{a^{2}} are the roots of the given equation. \end{array}$

Answer:

We write, $- 9 (a + b) x = - 3 (2 a + b) x - 3 (a + 2 b) x$ as $9 x^{2} \times (2 a^{2} + 5 a b + 2 b^{2}) = 9 (2 a^{2} + 5 a b + 2 b^{2}) x^{2} = [- 3 (2 a + b) x] \times [- 3 (a + 2 b) x]$
$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0 \Rightarrow 9 x^{2} - 3 (2 a + b) x - 3 (a + 2 b) x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x [3 x - (2 a + b)] - (a + 2 b) [3 x - (2 a + b)] = 0 \Rightarrow [3 x - (2 a + b)] [3 x - (a + 2 b)] = 0$
$\Rightarrow 3 x - (2 a + b) = 0 or 3 x - (a + 2 b) = 0 \Rightarrow x = \frac{2 a + b}{3} or x = \frac{a + 2 b}{3}$
Hence, $\frac{2 a + b}{3}$ and $\frac{a + 2 b}{3}$ are the roots of the given equation.

Question 52:

We write, $- 9 (a + b) x = - 3 (2 a + b) x - 3 (a + 2 b) x$ as $9 x^{2} \times (2 a^{2} + 5 a b + 2 b^{2}) = 9 (2 a^{2} + 5 a b + 2 b^{2}) x^{2} = [- 3 (2 a + b) x] \times [- 3 (a + 2 b) x]$
$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0 \Rightarrow 9 x^{2} - 3 (2 a + b) x - 3 (a + 2 b) x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x [3 x - (2 a + b)] - (a + 2 b) [3 x - (2 a + b)] = 0 \Rightarrow [3 x - (2 a + b)] [3 x - (a + 2 b)] = 0$
$\Rightarrow 3 x - (2 a + b) = 0 or 3 x - (a + 2 b) = 0 \Rightarrow x = \frac{2 a + b}{3} or x = \frac{a + 2 b}{3}$
Hence, $\frac{2 a + b}{3}$ and $\frac{a + 2 b}{3}$ are the roots of the given equation.

Answer:

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 \Rightarrow \frac{16}{x} - \frac{15}{x + 1} = 1 \Rightarrow \frac{16 x + 16 - 15 x}{x (x + 1)} = 1 \Rightarrow \frac{x + 16}{x^{2} + x} = 1$
$\Rightarrow x^{2} + x = x + 16 (Cross multiplication) \Rightarrow x^{2} - 16 = 0 \Rightarrow (x + 4) (x - 4) = 0 \Rightarrow x + 4 = 0 or x - 4 = 0$
$\Rightarrow x = - 4 or x = 4$
Hence, −4 and 4 are the roots of the given equation.

Question 53:

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 \Rightarrow \frac{16}{x} - \frac{15}{x + 1} = 1 \Rightarrow \frac{16 x + 16 - 15 x}{x (x + 1)} = 1 \Rightarrow \frac{x + 16}{x^{2} + x} = 1$
$\Rightarrow x^{2} + x = x + 16 (Cross multiplication) \Rightarrow x^{2} - 16 = 0 \Rightarrow (x + 4) (x - 4) = 0 \Rightarrow x + 4 = 0 or x - 4 = 0$
$\Rightarrow x = - 4 or x = 4$
Hence, −4 and 4 are the roots of the given equation.

Answer:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2} \Rightarrow \frac{4}{x} - \frac{5}{2 x + 3} = 3 \Rightarrow \frac{8 x + 12 - 5 x}{x (2 x + 3)} = 3 \Rightarrow \frac{3 x + 12}{2 x^{2} + 3 x} = 3$
$\Rightarrow \frac{x + 4}{2 x^{2} + 3 x} = 1 \Rightarrow 2 x^{2} + 3 x = x + 4 (Cross multiplication) \Rightarrow 2 x^{2} + 2 x - 4 = 0 \Rightarrow x^{2} + x - 2 = 0$
$\Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 or x - 1 = 0$
$\Rightarrow x = - 2 or x = 1$
Hence, −2 and 1 are the roots of the given equation.

Question 54:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2} \Rightarrow \frac{4}{x} - \frac{5}{2 x + 3} = 3 \Rightarrow \frac{8 x + 12 - 5 x}{x (2 x + 3)} = 3 \Rightarrow \frac{3 x + 12}{2 x^{2} + 3 x} = 3$
$\Rightarrow \frac{x + 4}{2 x^{2} + 3 x} = 1 \Rightarrow 2 x^{2} + 3 x = x + 4 (Cross multiplication) \Rightarrow 2 x^{2} + 2 x - 4 = 0 \Rightarrow x^{2} + x - 2 = 0$
$\Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 or x - 1 = 0$
$\Rightarrow x = - 2 or x = 1$
Hence, −2 and 1 are the roots of the given equation.

Answer:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3} \Rightarrow \frac{3}{x + 1} - \frac{2}{3 x - 1} = \frac{1}{2} \Rightarrow \frac{9 x - 3 - 2 x - 2}{(x + 1) (3 x - 1)} = \frac{1}{2} \Rightarrow \frac{7 x - 5}{3 x^{2} + 2 x - 1} = \frac{1}{2}$
$\Rightarrow 3 x^{2} + 2 x - 1 = 14 x - 10 (Cross multiplication) \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow x^{2} - 3 x - x + 3 = 0$
$\Rightarrow x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x - 3 = 0 or x - 1 = 0 \Rightarrow x = 3 or x = 1$
Hence, 1 and 3 are the roots of the given equation.

Question 55:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3} \Rightarrow \frac{3}{x + 1} - \frac{2}{3 x - 1} = \frac{1}{2} \Rightarrow \frac{9 x - 3 - 2 x - 2}{(x + 1) (3 x - 1)} = \frac{1}{2} \Rightarrow \frac{7 x - 5}{3 x^{2} + 2 x - 1} = \frac{1}{2}$
$\Rightarrow 3 x^{2} + 2 x - 1 = 14 x - 10 (Cross multiplication) \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow x^{2} - 3 x - x + 3 = 0$
$\Rightarrow x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x - 3 = 0 or x - 1 = 0 \Rightarrow x = 3 or x = 1$
Hence, 1 and 3 are the roots of the given equation.

Answer:

(i)
$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5 \Rightarrow \frac{x + 5 - x + 1}{(x - 1) (x + 5)} = \frac{6}{7} \Rightarrow \frac{6}{x^{2} + 4 x - 5} = \frac{6}{7} \Rightarrow x^{2} + 4 x - 5 = 7$
$\Rightarrow x^{2} + 4 x - 12 = 0 \Rightarrow x^{2} + 6 x - 2 x - 12 = 0 \Rightarrow x (x + 6) - 2 (x + 6) = 0 \Rightarrow (x + 6) (x - 2) = 0$
$\Rightarrow x + 6 = 0 or x - 2 = 0 \Rightarrow x = - 6 or x = 2$
Hence, −6 and 2 are the roots of the given equation.

(ii)
$\frac{1}{2 x - 3} + \frac{1}{x - 5} = 1 \frac{1}{9} \Rightarrow \frac{1}{2 x - 3} + \frac{1}{x - 5} = \frac{10}{9} \Rightarrow \frac{(x - 5) + (2 x - 3)}{(2 x - 3) (x - 5)} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 3 x - 10 x + 15} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 13 x + 15} = \frac{10}{9} \Rightarrow 27 x - 72 = 20 x^{2} - 130 x + 150 \Rightarrow 20 x^{2} - 157 x + 222 = 0 \Rightarrow 20 x^{2} - 120 x - 37 x + 222 = 0 \Rightarrow 20 x (x - 6) - 37 (x - 6) = 0 \Rightarrow (20 x - 37) (x - 6) = 0 \Rightarrow 20 x - 37 = 0 or x - 6 = 0 \Rightarrow x = \frac{37}{20} or x = 6$

Question 56:

(i)
$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5 \Rightarrow \frac{x + 5 - x + 1}{(x - 1) (x + 5)} = \frac{6}{7} \Rightarrow \frac{6}{x^{2} + 4 x - 5} = \frac{6}{7} \Rightarrow x^{2} + 4 x - 5 = 7$
$\Rightarrow x^{2} + 4 x - 12 = 0 \Rightarrow x^{2} + 6 x - 2 x - 12 = 0 \Rightarrow x (x + 6) - 2 (x + 6) = 0 \Rightarrow (x + 6) (x - 2) = 0$
$\Rightarrow x + 6 = 0 or x - 2 = 0 \Rightarrow x = - 6 or x = 2$
Hence, −6 and 2 are the roots of the given equation.

(ii)
$\frac{1}{2 x - 3} + \frac{1}{x - 5} = 1 \frac{1}{9} \Rightarrow \frac{1}{2 x - 3} + \frac{1}{x - 5} = \frac{10}{9} \Rightarrow \frac{(x - 5) + (2 x - 3)}{(2 x - 3) (x - 5)} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 3 x - 10 x + 15} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 13 x + 15} = \frac{10}{9} \Rightarrow 27 x - 72 = 20 x^{2} - 130 x + 150 \Rightarrow 20 x^{2} - 157 x + 222 = 0 \Rightarrow 20 x^{2} - 120 x - 37 x + 222 = 0 \Rightarrow 20 x (x - 6) - 37 (x - 6) = 0 \Rightarrow (20 x - 37) (x - 6) = 0 \Rightarrow 20 x - 37 = 0 or x - 6 = 0 \Rightarrow x = \frac{37}{20} or x = 6$

Answer:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 x} = \frac{1}{2 a} + \frac{1}{b} \Rightarrow \frac{2 x - 2 a - b - 2 x}{2 x (2 a + b + 2 x)} = \frac{2 a + b}{2 a b} \Rightarrow \frac{- (2 a + b)}{4 x^{2} + 4 a x + 2 b x} = \frac{2 a + b}{2 a b}$
$\Rightarrow 4 x^{2} + 4 a x + 2 b x = - 2 a b \Rightarrow 4 x^{2} + 4 a x + 2 b x + 2 a b = 0 \Rightarrow 4 x (x + a) + 2 b (x + a) = 0 \Rightarrow (x + a) (4 x + 2 b) = 0$
$\Rightarrow x + a = 0 or 4 x + 2 b = 0 \Rightarrow x = - a or x = - \frac{b}{2}$
Hence, $- a$ and $- \frac{b}{2}$ are the roots of the given equation.

Question 57:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 x} = \frac{1}{2 a} + \frac{1}{b} \Rightarrow \frac{2 x - 2 a - b - 2 x}{2 x (2 a + b + 2 x)} = \frac{2 a + b}{2 a b} \Rightarrow \frac{- (2 a + b)}{4 x^{2} + 4 a x + 2 b x} = \frac{2 a + b}{2 a b}$
$\Rightarrow 4 x^{2} + 4 a x + 2 b x = - 2 a b \Rightarrow 4 x^{2} + 4 a x + 2 b x + 2 a b = 0 \Rightarrow 4 x (x + a) + 2 b (x + a) = 0 \Rightarrow (x + a) (4 x + 2 b) = 0$
$\Rightarrow x + a = 0 or 4 x + 2 b = 0 \Rightarrow x = - a or x = - \frac{b}{2}$
Hence, $- a$ and $- \frac{b}{2}$ are the roots of the given equation.

Answer:

$Given : \frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4 \frac{1}{4} \Rightarrow \frac{(x + 3)}{(x - 2)} - \frac{(1 - x)}{x} = \frac{17}{4} \Rightarrow \frac{x (x + 3) - (1 - x) (x - 2)}{(x - 2) x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x - (x - 2 - x^{2} + 2 x)}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x + x^{2} - 3 x + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow 8 x^{2} + 8 = 17 x^{2} - 34 x [On cross multiplying] \Rightarrow - 9 x^{2} + 34 x + 8 = 0 \Rightarrow 9 x^{2} - 34 x - 8 = 0 \Rightarrow 9 x^{2} - 36 x + 2 x - 8 = 0 \Rightarrow 9 x (x - 4) + 2 (x - 4) = 0 \Rightarrow (x - 4) (9 x + 2) = 0 \Rightarrow x - 4 = 0 or 9 x + 2 = 0 \Rightarrow x = 4 or x = \frac{- 2}{9} H ence, the roots of the equation are 4 and \frac{- 2}{9} .$

Question 58:

$Given : \frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4 \frac{1}{4} \Rightarrow \frac{(x + 3)}{(x - 2)} - \frac{(1 - x)}{x} = \frac{17}{4} \Rightarrow \frac{x (x + 3) - (1 - x) (x - 2)}{(x - 2) x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x - (x - 2 - x^{2} + 2 x)}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x + x^{2} - 3 x + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow 8 x^{2} + 8 = 17 x^{2} - 34 x [On cross multiplying] \Rightarrow - 9 x^{2} + 34 x + 8 = 0 \Rightarrow 9 x^{2} - 34 x - 8 = 0 \Rightarrow 9 x^{2} - 36 x + 2 x - 8 = 0 \Rightarrow 9 x (x - 4) + 2 (x - 4) = 0 \Rightarrow (x - 4) (9 x + 2) = 0 \Rightarrow x - 4 = 0 or 9 x + 2 = 0 \Rightarrow x = 4 or x = \frac{- 2}{9} H ence, the roots of the equation are 4 and \frac{- 2}{9} .$

Answer:

$\frac{3 x - 4}{7} + \frac{7}{3 x - 4} = \frac{5}{2}, x \neq \frac{4}{3} \Rightarrow \frac{{(3 x - 4)}^{2} + 49}{7 (3 x - 4)} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 16 + 49}{21 x - 28} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 65}{21 x - 28} = \frac{5}{2}$
$\Rightarrow 18 x^{2} - 48 x + 130 = 105 x - 140 \Rightarrow 18 x^{2} - 153 x + 270 = 0 \Rightarrow 2 x^{2} - 17 x + 30 = 0 \Rightarrow 2 x^{2} - 12 x - 5 x + 30 = 0$
$\Rightarrow 2 x (x - 6) - 5 (x - 6) = 0 \Rightarrow (x - 6) (2 x - 5) = 0 \Rightarrow x - 6 = 0 or 2 x - 5 = 0 \Rightarrow x = 6 or x = \frac{5}{2}$
Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

Question 59:

$\frac{3 x - 4}{7} + \frac{7}{3 x - 4} = \frac{5}{2}, x \neq \frac{4}{3} \Rightarrow \frac{{(3 x - 4)}^{2} + 49}{7 (3 x - 4)} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 16 + 49}{21 x - 28} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 65}{21 x - 28} = \frac{5}{2}$
$\Rightarrow 18 x^{2} - 48 x + 130 = 105 x - 140 \Rightarrow 18 x^{2} - 153 x + 270 = 0 \Rightarrow 2 x^{2} - 17 x + 30 = 0 \Rightarrow 2 x^{2} - 12 x - 5 x + 30 = 0$
$\Rightarrow 2 x (x - 6) - 5 (x - 6) = 0 \Rightarrow (x - 6) (2 x - 5) = 0 \Rightarrow x - 6 = 0 or 2 x - 5 = 0 \Rightarrow x = 6 or x = \frac{5}{2}$
Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

Answer:

(i)
$\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1 \Rightarrow \frac{x^{2} + {(x - 1)}^{2}}{x (x - 1)} = \frac{17}{4} \Rightarrow \frac{x^{2} + x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4}$
$\Rightarrow 8 x^{2} - 8 x + 4 = 17 x^{2} - 17 x \Rightarrow 9 x^{2} - 9 x - 4 = 0 \Rightarrow 9 x^{2} - 12 x + 3 x - 4 = 0 \Rightarrow 3 x (3 x - 4) + 1 (3 x - 4) = 0$
$\Rightarrow (3 x - 4) (3 x + 1) = 0 \Rightarrow 3 x - 4 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{4}{3} or x = - \frac{1}{3}$
Hence, $\frac{4}{3}$ and $- \frac{1}{3}$ are the roots of the given equation.

(ii)
$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2 \Rightarrow \frac{(x - 1)^{2} + (2 x + 1)^{2}}{(2 x + 1) (x - 1)} = 2 \Rightarrow (x^{2} + 1 - 2 x) + (4 x^{2} + 1 + 4 x) = 2 (2 x + 1) (x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 2 (2 x^{2} - x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 4 x^{2} - 2 x - 2 \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow x^{2} + 2 x + 2 x + 4 = 0 \Rightarrow x (x + 2) + 2 (x + 2) = 0 \Rightarrow (x + 2) (x + 2) = 0 \Rightarrow (x + 2) = 0 or (x + 2) = 0 \Rightarrow x = - 2 or x = - 2 \Rightarrow x = - 2$

Question 60:

(i)
$\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1 \Rightarrow \frac{x^{2} + {(x - 1)}^{2}}{x (x - 1)} = \frac{17}{4} \Rightarrow \frac{x^{2} + x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4}$
$\Rightarrow 8 x^{2} - 8 x + 4 = 17 x^{2} - 17 x \Rightarrow 9 x^{2} - 9 x - 4 = 0 \Rightarrow 9 x^{2} - 12 x + 3 x - 4 = 0 \Rightarrow 3 x (3 x - 4) + 1 (3 x - 4) = 0$
$\Rightarrow (3 x - 4) (3 x + 1) = 0 \Rightarrow 3 x - 4 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{4}{3} or x = - \frac{1}{3}$
Hence, $\frac{4}{3}$ and $- \frac{1}{3}$ are the roots of the given equation.

(ii)
$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2 \Rightarrow \frac{(x - 1)^{2} + (2 x + 1)^{2}}{(2 x + 1) (x - 1)} = 2 \Rightarrow (x^{2} + 1 - 2 x) + (4 x^{2} + 1 + 4 x) = 2 (2 x + 1) (x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 2 (2 x^{2} - x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 4 x^{2} - 2 x - 2 \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow x^{2} + 2 x + 2 x + 4 = 0 \Rightarrow x (x + 2) + 2 (x + 2) = 0 \Rightarrow (x + 2) (x + 2) = 0 \Rightarrow (x + 2) = 0 or (x + 2) = 0 \Rightarrow x = - 2 or x = - 2 \Rightarrow x = - 2$

Answer:

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1 \Rightarrow \frac{x^{2} + {(x + 1)}^{2}}{x (x + 1)} = \frac{34}{15} \Rightarrow \frac{x^{2} + x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15} \Rightarrow \frac{2 x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}$
$\Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x \Rightarrow 4 x^{2} + 4 x - 15 = 0 \Rightarrow 4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0$
$\Rightarrow (2 x + 5) (2 x - 3) = 0 \Rightarrow 2 x + 5 = 0 or 2 x - 3 = 0 \Rightarrow x = - \frac{5}{2} or x = \frac{3}{2}$
Hence, $- \frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

Question 61:

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1 \Rightarrow \frac{x^{2} + {(x + 1)}^{2}}{x (x + 1)} = \frac{34}{15} \Rightarrow \frac{x^{2} + x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15} \Rightarrow \frac{2 x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}$
$\Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x \Rightarrow 4 x^{2} + 4 x - 15 = 0 \Rightarrow 4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0$
$\Rightarrow (2 x + 5) (2 x - 3) = 0 \Rightarrow 2 x + 5 = 0 or 2 x - 3 = 0 \Rightarrow x = - \frac{5}{2} or x = \frac{3}{2}$
Hence, $- \frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

Answer:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = 3 \frac{1}{3}, x \neq 5, 7 \Rightarrow \frac{(x - 4) (x - 7) + (x - 5) (x - 6)}{(x - 5) (x - 7)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 11 x + 28 + x^{2} - 11 x + 30}{x^{2} - 12 x + 35} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 22 x + 58}{x^{2} - 12 x + 35} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 11 x + 29}{x^{2} - 12 x + 35} = \frac{5}{3} \Rightarrow 3 x^{2} - 33 x + 87 = 5 x^{2} - 60 x + 175 \Rightarrow 2 x^{2} - 27 x + 88 = 0$
$\Rightarrow 2 x^{2} - 16 x - 11 x + 88 = 0 \Rightarrow 2 x (x - 8) - 11 (x - 8) = 0 \Rightarrow (x - 8) (2 x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2 x - 11 = 0 \Rightarrow x = 8 or x = \frac{11}{2}$
Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

Question 62:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = 3 \frac{1}{3}, x \neq 5, 7 \Rightarrow \frac{(x - 4) (x - 7) + (x - 5) (x - 6)}{(x - 5) (x - 7)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 11 x + 28 + x^{2} - 11 x + 30}{x^{2} - 12 x + 35} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 22 x + 58}{x^{2} - 12 x + 35} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 11 x + 29}{x^{2} - 12 x + 35} = \frac{5}{3} \Rightarrow 3 x^{2} - 33 x + 87 = 5 x^{2} - 60 x + 175 \Rightarrow 2 x^{2} - 27 x + 88 = 0$
$\Rightarrow 2 x^{2} - 16 x - 11 x + 88 = 0 \Rightarrow 2 x (x - 8) - 11 (x - 8) = 0 \Rightarrow (x - 8) (2 x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2 x - 11 = 0 \Rightarrow x = 8 or x = \frac{11}{2}$
Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

Answer:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4 \Rightarrow \frac{(x - 1) (x - 4) + (x - 2) (x - 3)}{(x - 2) (x - 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 5 x + 4 + x^{2} - 5 x + 6}{x^{2} - 6 x + 8} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 10 x + 10}{x^{2} - 6 x + 8} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 5 x + 5}{x^{2} - 6 x + 8} = \frac{5}{3} \Rightarrow 3 x^{2} - 15 x + 15 = 5 x^{2} - 30 x + 40 \Rightarrow 2 x^{2} - 15 x + 25 = 0 \Rightarrow 2 x^{2} - 10 x - 5 x + 25 = 0$
$\Rightarrow 2 x (x - 5) - 5 (x - 5) = 0 \Rightarrow (x - 5) (2 x - 5) = 0 \Rightarrow x - 5 = 0 or 2 x - 5 = 0 \Rightarrow x = 5 or x = \frac{5}{2}$
Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.

Question 63:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4 \Rightarrow \frac{(x - 1) (x - 4) + (x - 2) (x - 3)}{(x - 2) (x - 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 5 x + 4 + x^{2} - 5 x + 6}{x^{2} - 6 x + 8} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 10 x + 10}{x^{2} - 6 x + 8} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 5 x + 5}{x^{2} - 6 x + 8} = \frac{5}{3} \Rightarrow 3 x^{2} - 15 x + 15 = 5 x^{2} - 30 x + 40 \Rightarrow 2 x^{2} - 15 x + 25 = 0 \Rightarrow 2 x^{2} - 10 x - 5 x + 25 = 0$
$\Rightarrow 2 x (x - 5) - 5 (x - 5) = 0 \Rightarrow (x - 5) (2 x - 5) = 0 \Rightarrow x - 5 = 0 or 2 x - 5 = 0 \Rightarrow x = 5 or x = \frac{5}{2}$
Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.

Answer:

$Given : \frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x} \Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 1) (x - 2)} = \frac{6}{x} \Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x} \Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12 [O n cross multiplying] \Rightarrow 3 x^{2} - 13 x + 12 = 0 \Rightarrow 3 x^{2} - (9 + 4) x + 12 = 0 \Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0 \Rightarrow 3 x (x - 3) - 4 (x - 3) = 0 \Rightarrow (3 x - 4) (x - 3) = 0 \Rightarrow 3 x - 4 = 0 or x - 3 = 0 \Rightarrow x = \frac{4}{3} or x = 3 H ence, the roots of the equation are \frac{4}{3} and 3.$

Question 64:

$Given : \frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x} \Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 1) (x - 2)} = \frac{6}{x} \Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x} \Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12 [O n cross multiplying] \Rightarrow 3 x^{2} - 13 x + 12 = 0 \Rightarrow 3 x^{2} - (9 + 4) x + 12 = 0 \Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0 \Rightarrow 3 x (x - 3) - 4 (x - 3) = 0 \Rightarrow (3 x - 4) (x - 3) = 0 \Rightarrow 3 x - 4 = 0 or x - 3 = 0 \Rightarrow x = \frac{4}{3} or x = 3 H ence, the roots of the equation are \frac{4}{3} and 3.$

Answer:

(i)

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4 \Rightarrow \frac{x + 2 + 2 x + 2}{(x + 1) (x + 2)} = \frac{5}{x + 4} \Rightarrow \frac{3 x + 4}{x^{2} + 3 x + 2} = \frac{5}{x + 4} \Rightarrow (3 x + 4) (x + 4) = 5 (x^{2} + 3 x + 2)$
$\Rightarrow 3 x^{2} + 16 x + 16 = 5 x^{2} + 15 x + 10 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0$
$\Rightarrow (x - 2) (2 x + 3) = 0 \Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, 2 and $- \frac{3}{2}$ are the roots of the given equation.

(ii)

$\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4} \frac{(5 x + 1) + (3 x + 3)}{(x + 1) (5 x + 1)} = \frac{5}{x + 4} (x + 4) [(5 x + 1) + (3 x + 3)] = 5 (x + 1) (5 x + 1) (x + 4) [8 x + 4] = 25 x^{2} + 30 x + 5 8 x^{2} + 4 x + 32 x + 16 = 25 x^{2} + 30 x + 5 8 x^{2} + 36 x + 16 = 25 x^{2} + 30 x + 5 17 x^{2} - 6 x - 11 = 0 17 x^{2} - 17 x + 11 x - 11 = 0 17 x (x - 1) + 11 (x - 1) = 0 (17 x + 11) (x - 1) = 0 (17 x + 11) = 0 or (x - 1) = 0 x = - \frac{11}{17} or x = 1$

Question 65:

(i)

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4 \Rightarrow \frac{x + 2 + 2 x + 2}{(x + 1) (x + 2)} = \frac{5}{x + 4} \Rightarrow \frac{3 x + 4}{x^{2} + 3 x + 2} = \frac{5}{x + 4} \Rightarrow (3 x + 4) (x + 4) = 5 (x^{2} + 3 x + 2)$
$\Rightarrow 3 x^{2} + 16 x + 16 = 5 x^{2} + 15 x + 10 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0$
$\Rightarrow (x - 2) (2 x + 3) = 0 \Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, 2 and $- \frac{3}{2}$ are the roots of the given equation.

(ii)

$\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4} \frac{(5 x + 1) + (3 x + 3)}{(x + 1) (5 x + 1)} = \frac{5}{x + 4} (x + 4) [(5 x + 1) + (3 x + 3)] = 5 (x + 1) (5 x + 1) (x + 4) [8 x + 4] = 25 x^{2} + 30 x + 5 8 x^{2} + 4 x + 32 x + 16 = 25 x^{2} + 30 x + 5 8 x^{2} + 36 x + 16 = 25 x^{2} + 30 x + 5 17 x^{2} - 6 x - 11 = 0 17 x^{2} - 17 x + 11 x - 11 = 0 17 x (x - 1) + 11 (x - 1) = 0 (17 x + 11) (x - 1) = 0 (17 x + 11) = 0 or (x - 1) = 0 x = - \frac{11}{17} or x = 1$

Answer:

$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2} \Rightarrow \frac{3 {(3 x - 1)}^{2} - 2 {(2 x + 3)}^{2}}{(2 x + 3) (3 x - 1)} = 5 \Rightarrow \frac{3 (9 x^{2} - 6 x + 1) - 2 (4 x^{2} + 12 x + 9)}{6 x^{2} + 7 x - 3} = 5 \Rightarrow \frac{27 x^{2} - 18 x + 3 - 8 x^{2} - 24 x - 18}{6 x^{2} + 7 x - 3} = 5$
$\Rightarrow \frac{19 x^{2} - 42 x - 15}{6 x^{2} + 7 x - 3} = 5 \Rightarrow 19 x^{2} - 42 x - 15 = 30 x^{2} + 35 x - 15 \Rightarrow 11 x^{2} + 77 x = 0 \Rightarrow 11 x (x + 7) = 0$
$\Rightarrow x = 0 or x + 7 = 0 \Rightarrow x = 0 or x = - 7$
Hence, 0 and −7 are the roots of the given equation.

Question 66:

$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2} \Rightarrow \frac{3 {(3 x - 1)}^{2} - 2 {(2 x + 3)}^{2}}{(2 x + 3) (3 x - 1)} = 5 \Rightarrow \frac{3 (9 x^{2} - 6 x + 1) - 2 (4 x^{2} + 12 x + 9)}{6 x^{2} + 7 x - 3} = 5 \Rightarrow \frac{27 x^{2} - 18 x + 3 - 8 x^{2} - 24 x - 18}{6 x^{2} + 7 x - 3} = 5$
$\Rightarrow \frac{19 x^{2} - 42 x - 15}{6 x^{2} + 7 x - 3} = 5 \Rightarrow 19 x^{2} - 42 x - 15 = 30 x^{2} + 35 x - 15 \Rightarrow 11 x^{2} + 77 x = 0 \Rightarrow 11 x (x + 7) = 0$
$\Rightarrow x = 0 or x + 7 = 0 \Rightarrow x = 0 or x = - 7$
Hence, 0 and −7 are the roots of the given equation.

Answer:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7} \Rightarrow \frac{3 {(7 x + 1)}^{2} - 4 {(5 x - 3)}^{2}}{(5 x - 3) (7 x + 1)} = 11 \Rightarrow \frac{3 (49 x^{2} + 14 x + 1) - 4 (25 x^{2} - 30 x + 9)}{35 x^{2} - 16 x - 3} = 11 \Rightarrow \frac{147 x^{2} + 42 x + 3 - 100 x^{2} + 120 x - 36}{35 x^{2} - 16 x - 3} = 11$
$\Rightarrow \frac{47 x^{2} + 162 x - 33}{35 x^{2} - 16 x - 3} = 11 \Rightarrow 47 x^{2} + 162 x - 33 = 385 x^{2} - 176 x - 33 \Rightarrow 338 x^{2} - 338 x = 0 \Rightarrow 338 x (x - 1) = 0$
$\Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$
Hence, 0 and 1 are the roots of the given equation.

Question 67:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7} \Rightarrow \frac{3 {(7 x + 1)}^{2} - 4 {(5 x - 3)}^{2}}{(5 x - 3) (7 x + 1)} = 11 \Rightarrow \frac{3 (49 x^{2} + 14 x + 1) - 4 (25 x^{2} - 30 x + 9)}{35 x^{2} - 16 x - 3} = 11 \Rightarrow \frac{147 x^{2} + 42 x + 3 - 100 x^{2} + 120 x - 36}{35 x^{2} - 16 x - 3} = 11$
$\Rightarrow \frac{47 x^{2} + 162 x - 33}{35 x^{2} - 16 x - 3} = 11 \Rightarrow 47 x^{2} + 162 x - 33 = 385 x^{2} - 176 x - 33 \Rightarrow 338 x^{2} - 338 x = 0 \Rightarrow 338 x (x - 1) = 0$
$\Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$
Hence, 0 and 1 are the roots of the given equation.

Answer:

$Given : (\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3 P utting \frac{4 x - 3}{2 x + 1} = y, we get: y - \frac{10}{y} = 3 \Rightarrow \frac{y^{2} - 10}{y} = 3 \Rightarrow y^{2} - 10 = 3 y [On cross multiplying] \Rightarrow y^{2} - 3 y - 10 = 0 \Rightarrow y^{2} - (5 - 2) y - 10 = 0 \Rightarrow y^{2} - 5 y + 2 y - 10 = 0 \Rightarrow y (y - 5) + 2 (y - 5) = 0 \Rightarrow (y - 5) (y + 2) = 0 \Rightarrow y - 5 = 0 or y + 2 = 0 \Rightarrow y = 5 or y = - 2 Case I If y = 5, we get : \frac{4 x - 3}{2 x + 1} = 5 \Rightarrow 4 x - 3 = 5 (2 x + 1) [On cross multiplying] \Rightarrow 4 x - 3 = 10 x + 5 \Rightarrow - 6 x = 8 \Rightarrow - 6 x = 8 \Rightarrow x = - \frac{8^{4}}{6^{3}} \Rightarrow x = - \frac{4}{3} Case II If y = - 2, we get : \frac{4 x - 3}{2 x + 1} = - 2 \begin{array}{l} \Rightarrow 4 x - 3 = - 2 (2 x + 1) \\ \Rightarrow 4 x - 3 = - 4 x - 2 \Rightarrow 8 x = 1 \Rightarrow x = \frac{1}{8} Hence, the roots of the equation are - \frac{4}{3} and \frac{1}{8} . \end{array}$

Question 68:

$Given : (\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3 P utting \frac{4 x - 3}{2 x + 1} = y, we get: y - \frac{10}{y} = 3 \Rightarrow \frac{y^{2} - 10}{y} = 3 \Rightarrow y^{2} - 10 = 3 y [On cross multiplying] \Rightarrow y^{2} - 3 y - 10 = 0 \Rightarrow y^{2} - (5 - 2) y - 10 = 0 \Rightarrow y^{2} - 5 y + 2 y - 10 = 0 \Rightarrow y (y - 5) + 2 (y - 5) = 0 \Rightarrow (y - 5) (y + 2) = 0 \Rightarrow y - 5 = 0 or y + 2 = 0 \Rightarrow y = 5 or y = - 2 Case I If y = 5, we get : \frac{4 x - 3}{2 x + 1} = 5 \Rightarrow 4 x - 3 = 5 (2 x + 1) [On cross multiplying] \Rightarrow 4 x - 3 = 10 x + 5 \Rightarrow - 6 x = 8 \Rightarrow - 6 x = 8 \Rightarrow x = - \frac{8^{4}}{6^{3}} \Rightarrow x = - \frac{4}{3} Case II If y = - 2, we get : \frac{4 x - 3}{2 x + 1} = - 2 \begin{array}{l} \Rightarrow 4 x - 3 = - 2 (2 x + 1) \\ \Rightarrow 4 x - 3 = - 4 x - 2 \Rightarrow 8 x = 1 \Rightarrow x = \frac{1}{8} Hence, the roots of the equation are - \frac{4}{3} and \frac{1}{8} . \end{array}$

Answer:

$Given : {(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0 P utting \frac{x}{x + 1} = y, we get : y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - (3 + 2) y + 6 = 0 \Rightarrow y^{2} - 3 y - 2 y + 6 = 0 \Rightarrow y (y - 3) - 2 (y - 3) = 0 \Rightarrow (y - 3) (y - 2) = 0 \Rightarrow y - 3 = 0 or y - 2 = 0 \Rightarrow y = 3 or y = 2 Case I If y = 3, w e g e t : \frac{x}{x + 1} = 3 \Rightarrow x = 3 (x + 1) [On cross multiplying] \Rightarrow x = 3 x + 3 \Rightarrow x = \frac{- 3}{2} Case II If y = 2, we get : \frac{x}{x + 1} = 2 \Rightarrow x = 2 (x + 1) \Rightarrow x = 2 x + 2 \Rightarrow - x = 2 \Rightarrow x = - 2 Hence, the roots of the equation are \frac{- 3}{2} and - 2 .$

Question 69:

$Given : {(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0 P utting \frac{x}{x + 1} = y, we get : y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - (3 + 2) y + 6 = 0 \Rightarrow y^{2} - 3 y - 2 y + 6 = 0 \Rightarrow y (y - 3) - 2 (y - 3) = 0 \Rightarrow (y - 3) (y - 2) = 0 \Rightarrow y - 3 = 0 or y - 2 = 0 \Rightarrow y = 3 or y = 2 Case I If y = 3, w e g e t : \frac{x}{x + 1} = 3 \Rightarrow x = 3 (x + 1) [On cross multiplying] \Rightarrow x = 3 x + 3 \Rightarrow x = \frac{- 3}{2} Case II If y = 2, we get : \frac{x}{x + 1} = 2 \Rightarrow x = 2 (x + 1) \Rightarrow x = 2 x + 2 \Rightarrow - x = 2 \Rightarrow x = - 2 Hence, the roots of the equation are \frac{- 3}{2} and - 2 .$

Answer:

$\begin{array}{l} \frac{a}{(x - b)} + \frac{b}{(x - a)} = 2 \\ \begin{array}{l} \Rightarrow [\frac{a}{(x - b)} - 1] + [\frac{b}{(x - a)} - 1] = 0 \\ \Rightarrow \frac{a - (x - b)}{x - b} + \frac{b - (x - a)}{x - a} = 0 \end{array} \end{array} \Rightarrow \frac{a - x + b}{x - b} + \frac{a - x + b}{x - a} = 0 \Rightarrow (a - x + b) [\frac{1}{(x - b)} + \frac{1}{(x - a)}] = 0 \Rightarrow (a - x + b) [\frac{(x - a) + (x - b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [\frac{2 x - (a + b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [2 x - (a + b)] = 0 \Rightarrow a - x + b = 0 or 2 x - (a + b) = 0 \Rightarrow x = a + b or x = \frac{a + b}{2} Hence, the roots of the equation are (a + b) and (\frac{a + b}{2}) .$

Question 70:

$\begin{array}{l} \frac{a}{(x - b)} + \frac{b}{(x - a)} = 2 \\ \begin{array}{l} \Rightarrow [\frac{a}{(x - b)} - 1] + [\frac{b}{(x - a)} - 1] = 0 \\ \Rightarrow \frac{a - (x - b)}{x - b} + \frac{b - (x - a)}{x - a} = 0 \end{array} \end{array} \Rightarrow \frac{a - x + b}{x - b} + \frac{a - x + b}{x - a} = 0 \Rightarrow (a - x + b) [\frac{1}{(x - b)} + \frac{1}{(x - a)}] = 0 \Rightarrow (a - x + b) [\frac{(x - a) + (x - b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [\frac{2 x - (a + b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [2 x - (a + b)] = 0 \Rightarrow a - x + b = 0 or 2 x - (a + b) = 0 \Rightarrow x = a + b or x = \frac{a + b}{2} Hence, the roots of the equation are (a + b) and (\frac{a + b}{2}) .$

Answer:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b) \Rightarrow [\frac{a}{(a x - 1)} - b] + [\frac{b}{(b x - 1)} - a] = 0 \Rightarrow \frac{a - b (a x - 1)}{a x - 1} + \frac{b - a (b x - 1)}{b x - 1} = 0 \Rightarrow \frac{a - a b x + b}{a x - 1} + \frac{a - a b x + b}{b x - 1} = 0 \Rightarrow (a - a b x + b) [\frac{1}{(a x - 1)} + \frac{1}{(b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(b x - 1) + (a x - 1)}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(a + b) x - 2}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [(a + b) x - 2] = 0 \Rightarrow a - a b x + b = 0 or (a + b) x - 2 = 0 \Rightarrow x = \frac{(a + b)}{a b} or x = \frac{2}{(a + b)} Hence, the roots of the equation are \frac{(a + b)}{a b} and \frac{2}{(a + b)} .$

Question 71:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b) \Rightarrow [\frac{a}{(a x - 1)} - b] + [\frac{b}{(b x - 1)} - a] = 0 \Rightarrow \frac{a - b (a x - 1)}{a x - 1} + \frac{b - a (b x - 1)}{b x - 1} = 0 \Rightarrow \frac{a - a b x + b}{a x - 1} + \frac{a - a b x + b}{b x - 1} = 0 \Rightarrow (a - a b x + b) [\frac{1}{(a x - 1)} + \frac{1}{(b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(b x - 1) + (a x - 1)}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(a + b) x - 2}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [(a + b) x - 2] = 0 \Rightarrow a - a b x + b = 0 or (a + b) x - 2 = 0 \Rightarrow x = \frac{(a + b)}{a b} or x = \frac{2}{(a + b)} Hence, the roots of the equation are \frac{(a + b)}{a b} and \frac{2}{(a + b)} .$

Answer:

$3^{(x + 2)} + 3^{- x} = 10 3^{x} . 9 + \frac{1}{3^{x}} = 10 Let 3^{x} be equal to y . ∴ 9 y + \frac{1}{y} = 10 \Rightarrow 9 y^{2} + 1 = 10 y \Rightarrow 9 y^{2} - 10 y + 1 = 0 \Rightarrow (y - 1) (9 y - 1) = 0 \Rightarrow y - 1 = 0 or 9 y - 1 = 0 \Rightarrow y = 1 or y = \frac{1}{9} \Rightarrow 3^{x} = 1 {or 3}^{x} = \frac{1}{9} \Rightarrow 3^{x} = 3^{0} {or 3}^{x} = {3^{-}}^{2} \Rightarrow x = 0 or x = - 2 Hence, 0 and - 2 are the roots of the given equation .$

Question 72:

$3^{(x + 2)} + 3^{- x} = 10 3^{x} . 9 + \frac{1}{3^{x}} = 10 Let 3^{x} be equal to y . ∴ 9 y + \frac{1}{y} = 10 \Rightarrow 9 y^{2} + 1 = 10 y \Rightarrow 9 y^{2} - 10 y + 1 = 0 \Rightarrow (y - 1) (9 y - 1) = 0 \Rightarrow y - 1 = 0 or 9 y - 1 = 0 \Rightarrow y = 1 or y = \frac{1}{9} \Rightarrow 3^{x} = 1 {or 3}^{x} = \frac{1}{9} \Rightarrow 3^{x} = 3^{0} {or 3}^{x} = {3^{-}}^{2} \Rightarrow x = 0 or x = - 2 Hence, 0 and - 2 are the roots of the given equation .$

Answer:

$Given : 4^{(x + 1)} + 4^{(1 - x)} = 10 \Rightarrow 4^{x} .4 + 4^{1} . \frac{1}{4^{x}} = 10 {Let 4}^{x} be y . \begin{array}{l} ∴ 4 y + \frac{4}{y} = 10 \\ \Rightarrow 4 y^{2} - 10 y + 4 = 0 \end{array} \Rightarrow 4 y^{2} - 8 y - 2 y + 4 = 0 \Rightarrow 4 y (y - 2) - 2 (y - 2) = 0 \begin{array}{l} \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \\ \Rightarrow 4^{x} = 2 or \frac{1}{2} \end{array} \Rightarrow 4^{x} = 2^{2 x} = 2^{1} or 2^{2 x} = 2^{- 1} \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{2} Hence, \frac{1}{2} and \frac{- 1}{2} are roots of the given equation .$

Question 73:

$Given : 4^{(x + 1)} + 4^{(1 - x)} = 10 \Rightarrow 4^{x} .4 + 4^{1} . \frac{1}{4^{x}} = 10 {Let 4}^{x} be y . \begin{array}{l} ∴ 4 y + \frac{4}{y} = 10 \\ \Rightarrow 4 y^{2} - 10 y + 4 = 0 \end{array} \Rightarrow 4 y^{2} - 8 y - 2 y + 4 = 0 \Rightarrow 4 y (y - 2) - 2 (y - 2) = 0 \begin{array}{l} \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \\ \Rightarrow 4^{x} = 2 or \frac{1}{2} \end{array} \Rightarrow 4^{x} = 2^{2 x} = 2^{1} or 2^{2 x} = 2^{- 1} \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{2} Hence, \frac{1}{2} and \frac{- 1}{2} are roots of the given equation .$

Answer:

$Given : 2^{2 x} - {3.2}^{(x + 2)} + 32 = 0 \Rightarrow {(2^{x})}^{2} - {3.2}^{x} {.2}^{2} + 32 = 0 {Let 2}^{x} be y . ∴ y^{2} - 12 y + 32 = 0 \Rightarrow y^{2} - 8 y - 4 y + 32 = 0 \Rightarrow y (y - 8) - 4 (y - 8) = 0 \Rightarrow (y - 8) = 0 or (y - 4) = 0 \Rightarrow y = 8 or y = 4 ∴ 2^{x} = 8 or 2^{x} = 4 \Rightarrow 2^{x} = 2^{3} or 2^{x} = 2^{2} \Rightarrow x = 2 or 3 Hence, 2 and 3 are the roots of the given equation .$

Question 1:

$Given : 2^{2 x} - {3.2}^{(x + 2)} + 32 = 0 \Rightarrow {(2^{x})}^{2} - {3.2}^{x} {.2}^{2} + 32 = 0 {Let 2}^{x} be y . ∴ y^{2} - 12 y + 32 = 0 \Rightarrow y^{2} - 8 y - 4 y + 32 = 0 \Rightarrow y (y - 8) - 4 (y - 8) = 0 \Rightarrow (y - 8) = 0 or (y - 4) = 0 \Rightarrow y = 8 or y = 4 ∴ 2^{x} = 8 or 2^{x} = 4 \Rightarrow 2^{x} = 2^{3} or 2^{x} = 2^{2} \Rightarrow x = 2 or 3 Hence, 2 and 3 are the roots of the given equation .$

Answer:

(i) 2 x^{2} - 7 x + 6 = 0 Here, a = 2, b = - 7, c = 6 Discriminant D is diven by : D = b^{2} - 4 a c = {(- 7)}^{2} - 4 \times 2 \times 6 = 49 - 48 = 1

(ii) 3 x^{2} - 2 x + 8 = 0 Here, a = 3, b = - 2, c = 8 Discriminant D is given by : D = b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 8 = 4 - 96 = - 92

â€‹

(iii) 2 x^{2} - 5 \sqrt{2 x} + 4 = 0 Here, a = 2, b = - 5 \sqrt{2}, c = 4 Discriminant D is given by : D = b^{2} - 4 a c = {(- 5 \sqrt{2})}^{2} - 4 \times 2 \times 4 = (25 \times 2) - 32 = 50 - 32 = 18

â€‹

(iv) \sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0 Here, a = \sqrt{3}, b = 2 \sqrt{2}, c = - 2 \sqrt{3} Discriminant D is given by : D = b^{2} - 4 a c = {(2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = (4 \times 2) + (8 \times 3) = 8 + 24 = 32

(v)

(x - 1) (2 x - 1) = 0

\Rightarrow 2 x^{2} - 3 x + 1 = 0

Comparing it with

a x^{2} + b x + c = 0

, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D =

b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1

â€‹

(vi) 1 - x = 2 x^{2} \Rightarrow 2 x^{2} + x - 1 = 0 Here, a = 2, b = 1, c = - 1 Discriminant D is given by : D = b^{2} - 4 a c {= 1}^{2} - 4 \times 2 (- 1) = 1 + 8 = 9

Question 2:

(i) 2 x^{2} - 7 x + 6 = 0 Here, a = 2, b = - 7, c = 6 Discriminant D is diven by : D = b^{2} - 4 a c = {(- 7)}^{2} - 4 \times 2 \times 6 = 49 - 48 = 1

(ii) 3 x^{2} - 2 x + 8 = 0 Here, a = 3, b = - 2, c = 8 Discriminant D is given by : D = b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 8 = 4 - 96 = - 92

â€‹

(iii) 2 x^{2} - 5 \sqrt{2 x} + 4 = 0 Here, a = 2, b = - 5 \sqrt{2}, c = 4 Discriminant D is given by : D = b^{2} - 4 a c = {(- 5 \sqrt{2})}^{2} - 4 \times 2 \times 4 = (25 \times 2) - 32 = 50 - 32 = 18

â€‹

(iv) \sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0 Here, a = \sqrt{3}, b = 2 \sqrt{2}, c = - 2 \sqrt{3} Discriminant D is given by : D = b^{2} - 4 a c = {(2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = (4 \times 2) + (8 \times 3) = 8 + 24 = 32

(v)

(x - 1) (2 x - 1) = 0

\Rightarrow 2 x^{2} - 3 x + 1 = 0

Comparing it with

a x^{2} + b x + c = 0

, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D =

b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1

â€‹

(vi) 1 - x = 2 x^{2} \Rightarrow 2 x^{2} + x - 1 = 0 Here, a = 2, b = 1, c = - 1 Discriminant D is given by : D = b^{2} - 4 a c {= 1}^{2} - 4 \times 2 (- 1) = 1 + 8 = 9

Answer:

$Given : x^{2} - 4 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0, we get : a = 1, b = - 4 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4)}^{2} - 4 \times 1 \times (- 1) = 16 + 4 = 20 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 4) + \sqrt{20}}{2 \times 1} = \frac{4 + 2 \sqrt{5}}{2} = \frac{2 (2 + \sqrt{5})}{2} = (2 + \sqrt{5}) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 4) - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = \frac{2 (2 - \sqrt{5})}{2} = (2 - \sqrt{5}) Thus, the roots of the equation are (2 + \sqrt{5}) and (2 - \sqrt{5}) .$

Question 3:

$Given : x^{2} - 4 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0, we get : a = 1, b = - 4 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4)}^{2} - 4 \times 1 \times (- 1) = 16 + 4 = 20 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 4) + \sqrt{20}}{2 \times 1} = \frac{4 + 2 \sqrt{5}}{2} = \frac{2 (2 + \sqrt{5})}{2} = (2 + \sqrt{5}) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 4) - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = \frac{2 (2 - \sqrt{5})}{2} = (2 - \sqrt{5}) Thus, the roots of the equation are (2 + \sqrt{5}) and (2 - \sqrt{5}) .$

Answer:

$Given : x^{2} - 6 x + 4 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 1, b = - 6 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 1 \times 4 = 36 - 16 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + \sqrt{20}}{2 \times 1} = \frac{6 + 2 \sqrt{5}}{2} = \frac{2 (3 + \sqrt{5})}{2} = (3 + \sqrt{5}) \begin{array}{l} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - \sqrt{20}}{2 \times 1} = \frac{6 - 2 \sqrt{5}}{2} = \frac{2 (3 - \sqrt{5})}{2} = (3 - \sqrt{5}) \\ Thus, the roots of the equation are (3 + 2 \sqrt{5}) and (3 - 2 \sqrt{5}) . \end{array}$

Question 4:

$Given : x^{2} - 6 x + 4 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 1, b = - 6 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 1 \times 4 = 36 - 16 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + \sqrt{20}}{2 \times 1} = \frac{6 + 2 \sqrt{5}}{2} = \frac{2 (3 + \sqrt{5})}{2} = (3 + \sqrt{5}) \begin{array}{l} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - \sqrt{20}}{2 \times 1} = \frac{6 - 2 \sqrt{5}}{2} = \frac{2 (3 - \sqrt{5})}{2} = (3 - \sqrt{5}) \\ Thus, the roots of the equation are (3 + 2 \sqrt{5}) and (3 - 2 \sqrt{5}) . \end{array}$

Answer:

The given equation is $2 x^{2} + x - 4 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = $b^{2} - 4 a c = {(1)}^{2} - 4 \times 2 \times (- 4) = 1 + 32 = 33 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{33}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 1 + \sqrt{33}}{2 \times 2} = \frac{- 1 + \sqrt{33}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 1 - \sqrt{33}}{2 \times 2} = \frac{- 1 - \sqrt{33}}{4}$

Hence, $\frac{- 1 + \sqrt{33}}{4}$ and $\frac{- 1 - \sqrt{33}}{4}$ are the roots of the given equation.

Question 5:

The given equation is $2 x^{2} + x - 4 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = $b^{2} - 4 a c = {(1)}^{2} - 4 \times 2 \times (- 4) = 1 + 32 = 33 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{33}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 1 + \sqrt{33}}{2 \times 2} = \frac{- 1 + \sqrt{33}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 1 - \sqrt{33}}{2 \times 2} = \frac{- 1 - \sqrt{33}}{4}$

Hence, $\frac{- 1 + \sqrt{33}}{4}$ and $\frac{- 1 - \sqrt{33}}{4}$ are the roots of the given equation.

Answer:

$Given: 25 x^{2} + 30 x + 7 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 25, b = 30 and c = 7 Discriminant D is given by : D = (b^{2} - 4 a c) {= 30}^{2} - 4 \times 25 \times 7 = 900 - 700 = 200 = 200 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 30 + \sqrt{200}}{2 \times 25} = \frac{- 30 + 10 \sqrt{2}}{50} = \frac{10 (- 3 + \sqrt{2})}{50} = \frac{(- 3 + \sqrt{2})}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 30 - \sqrt{200}}{2 \times 25} = \frac{- 30 - 10 \sqrt{2}}{50} = \frac{10 (- 3 - \sqrt{2})}{50} = \frac{(- 3 - \sqrt{2})}{5} Thus, the roots of the equation are \frac{(- 3 + \sqrt{2})}{5} and \frac{(- 3 - \sqrt{2})}{5} .$

Question 6:

$Given: 25 x^{2} + 30 x + 7 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 25, b = 30 and c = 7 Discriminant D is given by : D = (b^{2} - 4 a c) {= 30}^{2} - 4 \times 25 \times 7 = 900 - 700 = 200 = 200 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 30 + \sqrt{200}}{2 \times 25} = \frac{- 30 + 10 \sqrt{2}}{50} = \frac{10 (- 3 + \sqrt{2})}{50} = \frac{(- 3 + \sqrt{2})}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 30 - \sqrt{200}}{2 \times 25} = \frac{- 30 - 10 \sqrt{2}}{50} = \frac{10 (- 3 - \sqrt{2})}{50} = \frac{(- 3 - \sqrt{2})}{5} Thus, the roots of the equation are \frac{(- 3 + \sqrt{2})}{5} and \frac{(- 3 - \sqrt{2})}{5} .$

Answer:

$Given : 16 x^{2} = 24 x + 1 ⇒ 16 x^{2} - 24 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0 a = 16, b = - 24 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 24)}^{2} - 4 \times 16 \times (- 1) = 576 + (64) = 640 > 0 Hence, the roots of the equation are real, Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 24) + \sqrt{640}}{2 \times 16} = \frac{24 + 8 \sqrt{10}}{32} = \frac{8 (3 + \sqrt{10})}{32} = \frac{(3 + \sqrt{10})}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 24) - \sqrt{640}}{2 \times 16} = \frac{24 - 8 \sqrt{10}}{32} = \frac{8 (3 - \sqrt{10})}{32} = \frac{(3 - \sqrt{10})}{4} Thus, the roots of the equation are \frac{(3 + \sqrt{10})}{4} and \frac{(3 - \sqrt{10})}{4} .$

Question 7:

$Given : 16 x^{2} = 24 x + 1 ⇒ 16 x^{2} - 24 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0 a = 16, b = - 24 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 24)}^{2} - 4 \times 16 \times (- 1) = 576 + (64) = 640 > 0 Hence, the roots of the equation are real, Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 24) + \sqrt{640}}{2 \times 16} = \frac{24 + 8 \sqrt{10}}{32} = \frac{8 (3 + \sqrt{10})}{32} = \frac{(3 + \sqrt{10})}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 24) - \sqrt{640}}{2 \times 16} = \frac{24 - 8 \sqrt{10}}{32} = \frac{8 (3 - \sqrt{10})}{32} = \frac{(3 - \sqrt{10})}{4} Thus, the roots of the equation are \frac{(3 + \sqrt{10})}{4} and \frac{(3 - \sqrt{10})}{4} .$

Answer:

$Given: 15 x^{2} - 28 = x ⇒ 15 x^{2} - x - 28 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 15, b = - 1 and c = - 28 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 1)}^{2} - 4 \times 15 \times (- 28) = 1 - (- 1680) = 1 + 1680 = 1681 = 1681 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 1) + \sqrt{1681}}{2 \times 15} = \frac{1 + 41}{30} = \frac{42}{30} = \frac{7}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 1) - \sqrt{1681}}{2 \times 15} = \frac{1 - 41}{30} = \frac{- 40}{30} = \frac{- 4}{3} Thus, the roots of the equation are \frac{7}{5} and \frac{- 4}{3} .$

Question 8:

$Given: 15 x^{2} - 28 = x ⇒ 15 x^{2} - x - 28 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 15, b = - 1 and c = - 28 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 1)}^{2} - 4 \times 15 \times (- 28) = 1 - (- 1680) = 1 + 1680 = 1681 = 1681 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 1) + \sqrt{1681}}{2 \times 15} = \frac{1 + 41}{30} = \frac{42}{30} = \frac{7}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 1) - \sqrt{1681}}{2 \times 15} = \frac{1 - 41}{30} = \frac{- 40}{30} = \frac{- 4}{3} Thus, the roots of the equation are \frac{7}{5} and \frac{- 4}{3} .$

Answer:

The given equation is $2 x^{2} - 2 \sqrt{2} x + 1 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $- 2 \sqrt{2}$ and c = 1

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times 2 \times 1 = 8 - 8 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

Question 9:

The given equation is $2 x^{2} - 2 \sqrt{2} x + 1 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $- 2 \sqrt{2}$ and c = 1

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times 2 \times 1 = 8 - 8 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

Answer:

The given equation is $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{2}$ , b = 7 and c = $5 \sqrt{2}$

∴ Discriminant, D = $b^{2} - 4 a c = {(7)}^{2} - 4 \times \sqrt{2} \times 5 \sqrt{2} = 49 - 40 = 9 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9} = 3$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 7 + 3}{2 \times \sqrt{2}} = \frac{- 4}{2 \sqrt{2}} = - \sqrt{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 7 - 3}{2 \times \sqrt{2}} = \frac{- 10}{2 \sqrt{2}} = - \frac{5 \sqrt{2}}{2}$

Hence, $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ are the roots of the given equation.

Question 10:

The given equation is $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{2}$ , b = 7 and c = $5 \sqrt{2}$

∴ Discriminant, D = $b^{2} - 4 a c = {(7)}^{2} - 4 \times \sqrt{2} \times 5 \sqrt{2} = 49 - 40 = 9 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9} = 3$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 7 + 3}{2 \times \sqrt{2}} = \frac{- 4}{2 \sqrt{2}} = - \sqrt{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 7 - 3}{2 \times \sqrt{2}} = \frac{- 10}{2 \sqrt{2}} = - \frac{5 \sqrt{2}}{2}$

Hence, $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ are the roots of the given equation.

Answer:

$Given : \sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = \sqrt{3}, b = 10 and c = - 8 \sqrt{3} Discriminant D is given by : D = (b^{2} - 4 a c) = {(10)}^{2} - 4 \times \sqrt{3} \times (- 8 \sqrt{3}) = 100 + 96 = 196 > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 10 + \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 10 - \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 - 14}{2 \sqrt{3}} = \frac{- 24}{2 \sqrt{3}} = \frac{- 12}{\sqrt{3}} = \frac{- 12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{- 12 \sqrt{3}}{3} = - 4 \sqrt{3} Thus, the roots of the equation are \frac{2 \sqrt{3}}{3} and - 4 \sqrt{3} .$

Question 11:

$Given : \sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = \sqrt{3}, b = 10 and c = - 8 \sqrt{3} Discriminant D is given by : D = (b^{2} - 4 a c) = {(10)}^{2} - 4 \times \sqrt{3} \times (- 8 \sqrt{3}) = 100 + 96 = 196 > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 10 + \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 10 - \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 - 14}{2 \sqrt{3}} = \frac{- 24}{2 \sqrt{3}} = \frac{- 12}{\sqrt{3}} = \frac{- 12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{- 12 \sqrt{3}}{3} = - 4 \sqrt{3} Thus, the roots of the equation are \frac{2 \sqrt{3}}{3} and - 4 \sqrt{3} .$

Answer:

The given equation is $\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{3}$ , b = $- 2 \sqrt{2}$ and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 8 + 24 = 32 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{32} = 4 \sqrt{2}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{6 \sqrt{2}}{2 \sqrt{3}} = \sqrt{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{- 2 \sqrt{2}}{2 \sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

Question 12:

The given equation is $\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{3}$ , b = $- 2 \sqrt{2}$ and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 8 + 24 = 32 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{32} = 4 \sqrt{2}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{6 \sqrt{2}}{2 \sqrt{3}} = \sqrt{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{- 2 \sqrt{2}}{2 \sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

Answer:

The given equation is $2 x^{2} + 6 \sqrt{3} x - 60 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $6 \sqrt{3}$ and c = $- 60$

∴ Discriminant, D = $b^{2} - 4 a c = {(6 \sqrt{3})}^{2} - 4 \times 2 \times (- 60) = 108 + 480 = 588 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{588} = 14 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} + 14 \sqrt{3}}{2 \times 2} = \frac{8 \sqrt{3}}{4} = 2 \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} - 14 \sqrt{3}}{2 \times 2} = \frac{- 20 \sqrt{3}}{4} = - 5 \sqrt{3}$

Hence, $2 \sqrt{3}$ and $- 5 \sqrt{3}$ are the roots of the given equation.

Question 13:

The given equation is $2 x^{2} + 6 \sqrt{3} x - 60 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $6 \sqrt{3}$ and c = $- 60$

∴ Discriminant, D = $b^{2} - 4 a c = {(6 \sqrt{3})}^{2} - 4 \times 2 \times (- 60) = 108 + 480 = 588 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{588} = 14 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} + 14 \sqrt{3}}{2 \times 2} = \frac{8 \sqrt{3}}{4} = 2 \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} - 14 \sqrt{3}}{2 \times 2} = \frac{- 20 \sqrt{3}}{4} = - 5 \sqrt{3}$

Hence, $2 \sqrt{3}$ and $- 5 \sqrt{3}$ are the roots of the given equation.

Answer:

The given equation is $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $4 \sqrt{3}$ , b = 5 and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = 5^{2} - 4 \times 4 \sqrt{3} \times (- 2 \sqrt{3}) = 25 + 96 = 121 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{121} = 11$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 + 11}{2 \times 4 \sqrt{3}} = \frac{6}{8 \sqrt{3}} = \frac{\sqrt{3}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 - 11}{2 \times 4 \sqrt{3}} = \frac{- 16}{8 \sqrt{3}} = - \frac{2 \sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{4}$ and $- \frac{2 \sqrt{3}}{3}$ are the roots of the given equation.

Question 14:

The given equation is $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $4 \sqrt{3}$ , b = 5 and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = 5^{2} - 4 \times 4 \sqrt{3} \times (- 2 \sqrt{3}) = 25 + 96 = 121 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{121} = 11$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 + 11}{2 \times 4 \sqrt{3}} = \frac{6}{8 \sqrt{3}} = \frac{\sqrt{3}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 - 11}{2 \times 4 \sqrt{3}} = \frac{- 16}{8 \sqrt{3}} = - \frac{2 \sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{4}$ and $- \frac{2 \sqrt{3}}{3}$ are the roots of the given equation.

Answer:

The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = $- 2 \sqrt{6}$ and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) + 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) - 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

Question 15:

The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = $- 2 \sqrt{6}$ and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) + 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) - 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

Answer:

The given equation is $2 \sqrt{3} x^{2} - 5 x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $2 \sqrt{3}$ , b = $- 5$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 5)}^{2} - 4 \times 2 \sqrt{3} \times \sqrt{3} = 25 - 24 = 1 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{1} = 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 5) + 1}{2 \times 2 \sqrt{3}} = \frac{6}{4 \sqrt{3}} = \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 5) - 1}{2 \times 2 \sqrt{3}} = \frac{4}{4 \sqrt{3}} = \frac{\sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

Question 16:

The given equation is $2 \sqrt{3} x^{2} - 5 x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $2 \sqrt{3}$ , b = $- 5$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 5)}^{2} - 4 \times 2 \sqrt{3} \times \sqrt{3} = 25 - 24 = 1 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{1} = 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 5) + 1}{2 \times 2 \sqrt{3}} = \frac{6}{4 \sqrt{3}} = \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 5) - 1}{2 \times 2 \sqrt{3}} = \frac{4}{4 \sqrt{3}} = \frac{\sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

Answer:

The given equation is $x^{2} + x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = 1^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Question 17:

The given equation is $x^{2} + x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = 1^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Answer:

The given equation is $2 x^{2} + a x - a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 2, B = a and C = $- a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = a^{2} - 4 \times 2 \times - a^{2} = a^{2} + 8 a^{2} = 9 a^{2} \geq 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9 a^{2}} = 3 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- a + 3 a}{2 \times 2} = \frac{2 a}{4} = \frac{a}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- a - 3 a}{2 \times 2} = \frac{- 4 a}{4} = - a$

Hence, $\frac{a}{2}$ and $- a$ are the roots of the given equation.

Question 18:

The given equation is $2 x^{2} + a x - a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 2, B = a and C = $- a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = a^{2} - 4 \times 2 \times - a^{2} = a^{2} + 8 a^{2} = 9 a^{2} \geq 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9 a^{2}} = 3 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- a + 3 a}{2 \times 2} = \frac{2 a}{4} = \frac{a}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- a - 3 a}{2 \times 2} = \frac{- 4 a}{4} = - a$

Hence, $\frac{a}{2}$ and $- a$ are the roots of the given equation.

Answer:

The given equation is $x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = $- (\sqrt{3} + 1)$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {[- (\sqrt{3} + 1)]}^{2} - 4 \times 1 \times \sqrt{3} = 3 + 1 + 2 \sqrt{3} - 4 \sqrt{3} = 3 - 2 \sqrt{3} + 1 = {(\sqrt{3} - 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(\sqrt{3} - 1)}^{2}} = \sqrt{3} - 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] + (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] - (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2} = \frac{2}{2} = 1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

Question 19:

The given equation is $x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = $- (\sqrt{3} + 1)$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {[- (\sqrt{3} + 1)]}^{2} - 4 \times 1 \times \sqrt{3} = 3 + 1 + 2 \sqrt{3} - 4 \sqrt{3} = 3 - 2 \sqrt{3} + 1 = {(\sqrt{3} - 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(\sqrt{3} - 1)}^{2}} = \sqrt{3} - 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] + (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] - (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2} = \frac{2}{2} = 1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

Answer:

The given equation is $2 x^{2} + 5 \sqrt{3} x + 6 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $5 \sqrt{3}$ and c = 6

∴ Discriminant, D = $b^{2} - 4 a c = {(5 \sqrt{3})}^{2} - 4 \times 2 \times 6 = 75 - 48 = 27 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{27} = 3 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} + 3 \sqrt{3}}{2 \times 2} = \frac{- 2 \sqrt{3}}{4} = - \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} - 3 \sqrt{3}}{2 \times 2} = \frac{- 8 \sqrt{3}}{4} = - 2 \sqrt{3}$

Hence, $- \frac{\sqrt{3}}{2}$ and $- 2 \sqrt{3}$ are the roots of the given equation.

Question 20:

The given equation is $2 x^{2} + 5 \sqrt{3} x + 6 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $5 \sqrt{3}$ and c = 6

∴ Discriminant, D = $b^{2} - 4 a c = {(5 \sqrt{3})}^{2} - 4 \times 2 \times 6 = 75 - 48 = 27 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{27} = 3 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} + 3 \sqrt{3}}{2 \times 2} = \frac{- 2 \sqrt{3}}{4} = - \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} - 3 \sqrt{3}}{2 \times 2} = \frac{- 8 \sqrt{3}}{4} = - 2 \sqrt{3}$

Hence, $- \frac{\sqrt{3}}{2}$ and $- 2 \sqrt{3}$ are the roots of the given equation.

Answer:

The given equation is $3 x^{2} - 2 x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 2 = 4 - 24 = - 20 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Question 21:

The given equation is $3 x^{2} - 2 x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 2 = 4 - 24 = - 20 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Answer:

The given equation is

$x + \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} + 1}{x} = 3 \Rightarrow x^{2} + 1 = 3 x \Rightarrow x^{2} - 3 x + 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times 1 = 9 - 4 = 5 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{5}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{5}}{2 \times 1} = \frac{3 + \sqrt{5}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{5}}{2 \times 1} = \frac{3 - \sqrt{5}}{2}$

Hence, $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ are the roots of the given equation.

Question 22:

The given equation is

$x + \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} + 1}{x} = 3 \Rightarrow x^{2} + 1 = 3 x \Rightarrow x^{2} - 3 x + 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times 1 = 9 - 4 = 5 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{5}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{5}}{2 \times 1} = \frac{3 + \sqrt{5}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{5}}{2 \times 1} = \frac{3 - \sqrt{5}}{2}$

Hence, $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ are the roots of the given equation.

Answer:

The given equation is

$\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2 \Rightarrow \frac{x - 2 - x}{x (x - 2)} = 3 \Rightarrow \frac{- 2}{x^{2} - 2 x} = 3 \Rightarrow - 2 = 3 x^{2} - 6 x$

$\Rightarrow 3 x^{2} - 6 x + 2 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = −6 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 6)}^{2} - 4 \times 3 \times 2 = 36 - 24 = 12 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{12} = 2 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + 2 \sqrt{3}}{2 \times 3} = \frac{6 + 2 \sqrt{3}}{6} = \frac{3 + \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - 2 \sqrt{3}}{2 \times 3} = \frac{6 - 2 \sqrt{3}}{6} = \frac{3 - \sqrt{3}}{3}$

Hence, $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$ are the roots of the given equation.

Question 23:

The given equation is

$\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2 \Rightarrow \frac{x - 2 - x}{x (x - 2)} = 3 \Rightarrow \frac{- 2}{x^{2} - 2 x} = 3 \Rightarrow - 2 = 3 x^{2} - 6 x$

$\Rightarrow 3 x^{2} - 6 x + 2 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = −6 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 6)}^{2} - 4 \times 3 \times 2 = 36 - 24 = 12 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{12} = 2 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + 2 \sqrt{3}}{2 \times 3} = \frac{6 + 2 \sqrt{3}}{6} = \frac{3 + \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - 2 \sqrt{3}}{2 \times 3} = \frac{6 - 2 \sqrt{3}}{6} = \frac{3 - \sqrt{3}}{3}$

Hence, $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$ are the roots of the given equation.

Answer:

The given equation is

$x - \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} - 1}{x} = 3 \Rightarrow x^{2} - 1 = 3 x \Rightarrow x^{2} - 3 x - 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = −1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times (- 1) = 9 + 4 = 13 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{13}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{13}}{2 \times 1} = \frac{3 + \sqrt{13}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{13}}{2 \times 1} = \frac{3 - \sqrt{13}}{2}$

Hence, $\frac{3 + \sqrt{13}}{2}$ and $\frac{3 - \sqrt{13}}{2}$ are the roots of the given equation.

Question 24:

The given equation is

$x - \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} - 1}{x} = 3 \Rightarrow x^{2} - 1 = 3 x \Rightarrow x^{2} - 3 x - 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = −1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times (- 1) = 9 + 4 = 13 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{13}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{13}}{2 \times 1} = \frac{3 + \sqrt{13}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{13}}{2 \times 1} = \frac{3 - \sqrt{13}}{2}$

Hence, $\frac{3 + \sqrt{13}}{2}$ and $\frac{3 - \sqrt{13}}{2}$ are the roots of the given equation.

Answer:

The given equation is

$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x \Rightarrow \frac{m^{2} x^{2} + n^{2}}{m n} = 1 - 2 x \Rightarrow m^{2} x^{2} + n^{2} = m n - 2 m n x \Rightarrow m^{2} x^{2} + 2 m n x + n^{2} - m n = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = $m^{2}$ , b = 2mn and c = $n^{2} - m n$ .

∴ Discriminant, D = $b^{2} - 4 a c = {(2 m n)}^{2} - 4 \times m^{2} \times (n^{2} - m n) = 4 m^{2} n^{2} - 4 m^{2} n^{2} + 4 m^{3} n = 4 m^{3} n > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 m^{3} n} = 2 m \sqrt{m n}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 2 m n + 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{2 m (- n + \sqrt{m n})}{2 m^{2}} = \frac{- n + \sqrt{m n}}{m} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 2 m n - 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{- 2 m (n + \sqrt{m n})}{2 m^{2}} = \frac{- n - \sqrt{m n}}{m}$

Hence, $\frac{- n + \sqrt{m n}}{m}$ and $\frac{- n - \sqrt{m n}}{m}$ are the roots of the given equation.

Question 25:

The given equation is

$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x \Rightarrow \frac{m^{2} x^{2} + n^{2}}{m n} = 1 - 2 x \Rightarrow m^{2} x^{2} + n^{2} = m n - 2 m n x \Rightarrow m^{2} x^{2} + 2 m n x + n^{2} - m n = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = $m^{2}$ , b = 2mn and c = $n^{2} - m n$ .

∴ Discriminant, D = $b^{2} - 4 a c = {(2 m n)}^{2} - 4 \times m^{2} \times (n^{2} - m n) = 4 m^{2} n^{2} - 4 m^{2} n^{2} + 4 m^{3} n = 4 m^{3} n > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 m^{3} n} = 2 m \sqrt{m n}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 2 m n + 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{2 m (- n + \sqrt{m n})}{2 m^{2}} = \frac{- n + \sqrt{m n}}{m} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 2 m n - 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{- 2 m (n + \sqrt{m n})}{2 m^{2}} = \frac{- n - \sqrt{m n}}{m}$

Hence, $\frac{- n + \sqrt{m n}}{m}$ and $\frac{- n - \sqrt{m n}}{m}$ are the roots of the given equation.

Answer:

The given equation is $36 x^{2} - 12 a x + (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 36, B = $- 12 a$ and C = $a^{2} - b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 12 a)}^{2} - 4 \times 36 \times (a^{2} - b^{2}) = 144 a^{2} - 144 a^{2} + 144 b^{2} = 144 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{144 b^{2}} = 12 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 12 a) + 12 b}{2 \times 36} = \frac{12 (a + b)}{72} = \frac{a + b}{6} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 12 a) - 12 b}{2 \times 36} = \frac{12 (a - b)}{72} = \frac{a - b}{6}$

Hence, $\frac{a + b}{6}$ and $\frac{a - b}{6}$ are the roots of the given equation.

Question 26:

The given equation is $36 x^{2} - 12 a x + (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 36, B = $- 12 a$ and C = $a^{2} - b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 12 a)}^{2} - 4 \times 36 \times (a^{2} - b^{2}) = 144 a^{2} - 144 a^{2} + 144 b^{2} = 144 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{144 b^{2}} = 12 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 12 a) + 12 b}{2 \times 36} = \frac{12 (a + b)}{72} = \frac{a + b}{6} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 12 a) - 12 b}{2 \times 36} = \frac{12 (a - b)}{72} = \frac{a - b}{6}$

Hence, $\frac{a + b}{6}$ and $\frac{a - b}{6}$ are the roots of the given equation.

Answer:

$Given: x^{2} - 2 a x + (a^{2} - b^{2}) = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 1, B = - 2 a and C = (a^{2} - b^{2}) Discriminant D is given by : D = B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - b^{2}) = 4 a^{2} - 4 a^{2} + 4 b^{2} = 4 b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 a) + \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a + 2 b}{2} = \frac{2 (a + b)}{2} = (a + b) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 a) - \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a - 2 b}{2} = \frac{2 (a - b)}{2} = (a - b) Hence, the roots of the equation are (a + b) and (a - b) .$

Question 27:

$Given: x^{2} - 2 a x + (a^{2} - b^{2}) = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 1, B = - 2 a and C = (a^{2} - b^{2}) Discriminant D is given by : D = B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - b^{2}) = 4 a^{2} - 4 a^{2} + 4 b^{2} = 4 b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 a) + \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a + 2 b}{2} = \frac{2 (a + b)}{2} = (a + b) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 a) - \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a - 2 b}{2} = \frac{2 (a - b)}{2} = (a - b) Hence, the roots of the equation are (a + b) and (a - b) .$

Answer:

The given equation is $x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- 2 a$ and C = $- (4 b^{2} - a^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times [- (4 b^{2} - a^{2})] = 4 a^{2} + 16 b^{2} - 4 a^{2} = 16 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{2}} = 4 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 2 a) + 4 b}{2 \times 1} = \frac{2 (a + 2 b)}{2} = a + 2 b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 2 a) - 4 b}{2 \times 1} = \frac{2 (a - 2 b)}{2} = a - 2 b$

Hence, $a + 2 b$ and $a - 2 b$ are the roots of the given equation.

Question 28:

The given equation is $x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- 2 a$ and C = $- (4 b^{2} - a^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times [- (4 b^{2} - a^{2})] = 4 a^{2} + 16 b^{2} - 4 a^{2} = 16 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{2}} = 4 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 2 a) + 4 b}{2 \times 1} = \frac{2 (a + 2 b)}{2} = a + 2 b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 2 a) - 4 b}{2 \times 1} = \frac{2 (a - 2 b)}{2} = a - 2 b$

Hence, $a + 2 b$ and $a - 2 b$ are the roots of the given equation.

Answer:

The given equation is $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 6 and C = $- (a^{2} + 2 a - 8)$

∴ Discriminant, D = $B^{2} - 4 A C = 6^{2} - 4 \times 1 \times [- (a^{2} + 2 a - 8)] = 36 + 4 a^{2} + 8 a - 32 = 4 a^{2} + 8 a + 4 = 4 (a^{2} + 2 a + 1) = 4 {(a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 {(a + 1)}^{2}} = 2 (a + 1)$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 6 + 2 (a + 1)}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 6 - 2 (a + 1)}{2 \times 1} = \frac{- 2 a - 8}{2} = - a - 4 = - (a + 4)$

Hence, $(a - 2)$ and $- (a + 4)$ are the roots of the given equation.

Question 29:

The given equation is $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 6 and C = $- (a^{2} + 2 a - 8)$

∴ Discriminant, D = $B^{2} - 4 A C = 6^{2} - 4 \times 1 \times [- (a^{2} + 2 a - 8)] = 36 + 4 a^{2} + 8 a - 32 = 4 a^{2} + 8 a + 4 = 4 (a^{2} + 2 a + 1) = 4 {(a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 {(a + 1)}^{2}} = 2 (a + 1)$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 6 + 2 (a + 1)}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 6 - 2 (a + 1)}{2 \times 1} = \frac{- 2 a - 8}{2} = - a - 4 = - (a + 4)$

Hence, $(a - 2)$ and $- (a + 4)$ are the roots of the given equation.

Answer:

The given equation is $x^{2} + 5 x - (a^{2} + a - 6) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 5 and C = $- (a^{2} + a - 6)$

∴ Discriminant, D = $B^{2} - 4 A C = 5^{2} - 4 \times 1 \times [- (a^{2} + a - 6)] = 25 + 4 a^{2} + 4 a - 24 = 4 a^{2} + 4 a + 1 = {(2 a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(2 a + 1)}^{2}} = 2 a + 1$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 5 + 2 a + 1}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 5 - (2 a + 1)}{2 \times 1} = \frac{- 2 a - 6}{2} = - a - 3 = - (a + 3)$

Hence, $(a - 2)$ and $- (a + 3)$ are the roots of the given equation.

Question 30:

The given equation is $x^{2} + 5 x - (a^{2} + a - 6) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 5 and C = $- (a^{2} + a - 6)$

∴ Discriminant, D = $B^{2} - 4 A C = 5^{2} - 4 \times 1 \times [- (a^{2} + a - 6)] = 25 + 4 a^{2} + 4 a - 24 = 4 a^{2} + 4 a + 1 = {(2 a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(2 a + 1)}^{2}} = 2 a + 1$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 5 + 2 a + 1}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 5 - (2 a + 1)}{2 \times 1} = \frac{- 2 a - 6}{2} = - a - 3 = - (a + 3)$

Hence, $(a - 2)$ and $- (a + 3)$ are the roots of the given equation.

Answer:

The given equation is $x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = −4a and C = $- b^{2} + 4 a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a)}^{2} - 4 \times 1 \times (- b^{2} + 4 a^{2}) = 16 a^{2} + 4 b^{2} - 16 a^{2} = 4 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 b^{2}} = 2 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a) + 2 b}{2 \times 1} = \frac{4 a + 2 b}{2} = 2 a + b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a) - 2 b}{2 \times 1} = \frac{4 a - 2 b}{2} = 2 a - b$

Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Question 31:

The given equation is $x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = −4a and C = $- b^{2} + 4 a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a)}^{2} - 4 \times 1 \times (- b^{2} + 4 a^{2}) = 16 a^{2} + 4 b^{2} - 16 a^{2} = 4 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 b^{2}} = 2 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a) + 2 b}{2 \times 1} = \frac{4 a + 2 b}{2} = 2 a + b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a) - 2 b}{2 \times 1} = \frac{4 a - 2 b}{2} = 2 a - b$

Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Answer:

The given equation is $4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = −4a² and C = $a^{4} - b^{4}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a^{2})}^{2} - 4 \times 4 \times (a^{4} - b^{4}) = 16 a^{4} - 16 a^{4} + 16 b^{4} = 16 b^{4} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{4}} = 4 b^{2}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) + 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} + b^{2})}{8} = \frac{a^{2} + b^{2}}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) - 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} - b^{2})}{8} = \frac{a^{2} - b^{2}}{2}$

Hence, $\frac{1}{2} (a^{2} + b^{2})$ and $\frac{1}{2} (a^{2} - b^{2})$ are the roots of the given equation.

Question 32:

The given equation is $4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = −4a² and C = $a^{4} - b^{4}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a^{2})}^{2} - 4 \times 4 \times (a^{4} - b^{4}) = 16 a^{4} - 16 a^{4} + 16 b^{4} = 16 b^{4} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{4}} = 4 b^{2}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) + 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} + b^{2})}{8} = \frac{a^{2} + b^{2}}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) - 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} - b^{2})}{8} = \frac{a^{2} - b^{2}}{2}$

Hence, $\frac{1}{2} (a^{2} + b^{2})$ and $\frac{1}{2} (a^{2} - b^{2})$ are the roots of the given equation.

Answer:

The given equation is $4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = 4b and C = $- (a^{2} - b^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(4 b)}^{2} - 4 \times 4 \times [- (a^{2} - b^{2})] = 16 b^{2} + 16 a^{2} - 16 b^{2} = 16 a^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 a^{2}} = 4 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 4 b + 4 a}{2 \times 4} = \frac{4 (a - b)}{8} = \frac{a - b}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 4 b - 4 a}{2 \times 4} = \frac{- 4 (a + b)}{8} = - \frac{a + b}{2}$

Hence, $\frac{1}{2} (a - b)$ and $- \frac{1}{2} (a + b)$ are the roots of the given equation.

Question 33:

The given equation is $4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = 4b and C = $- (a^{2} - b^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(4 b)}^{2} - 4 \times 4 \times [- (a^{2} - b^{2})] = 16 b^{2} + 16 a^{2} - 16 b^{2} = 16 a^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 a^{2}} = 4 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 4 b + 4 a}{2 \times 4} = \frac{4 (a - b)}{8} = \frac{a - b}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 4 b - 4 a}{2 \times 4} = \frac{- 4 (a + b)}{8} = - \frac{a + b}{2}$

Hence, $\frac{1}{2} (a - b)$ and $- \frac{1}{2} (a + b)$ are the roots of the given equation.

Answer:

The given equation is $x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- (2 b - 1)$ and C = $b^{2} - b - 20$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (2 b - 1)]}^{2} - 4 \times 1 \times (b^{2} - b - 20) = 4 b^{2} - 4 b + 1 - 4 b^{2} + 4 b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{81} = 9$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] + 9}{2 \times 1} = \frac{2 b + 8}{2} = b + 4 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] - 9}{2 \times 1} = \frac{2 b - 10}{2} = b - 5$

Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.

Question 34:

The given equation is $x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- (2 b - 1)$ and C = $b^{2} - b - 20$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (2 b - 1)]}^{2} - 4 \times 1 \times (b^{2} - b - 20) = 4 b^{2} - 4 b + 1 - 4 b^{2} + 4 b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{81} = 9$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] + 9}{2 \times 1} = \frac{2 b + 8}{2} = b + 4 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] - 9}{2 \times 1} = \frac{2 b - 10}{2} = b - 5$

Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.

Answer:

$Given: {3a}^{2} x^{2} + 8 a b x + 4 b^{2} = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 3 a^{2}, B = 8 a b and C = 4 b^{2} Discriminant D is given by : D = (B^{2} - 4 A C) = {(8 a b)}^{2} - 4 \times 3 a^{2} \times 4 b^{2} {= 16a}^{2} b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 8 a b + \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b + 4 a b}{6 a^{2}} = \frac{- 4 a b}{6 a^{2}} = \frac{- 2 b}{3 a} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 8 a b - \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b - 4 a b}{6 a^{2}} = \frac{- 12 a b}{6 a^{2}} = \frac{- 2 b}{a} Thus, the roots of the equation are \frac{- 2 b}{3 a} and \frac{- 2 b}{a} .$

Question 35:

$Given: {3a}^{2} x^{2} + 8 a b x + 4 b^{2} = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 3 a^{2}, B = 8 a b and C = 4 b^{2} Discriminant D is given by : D = (B^{2} - 4 A C) = {(8 a b)}^{2} - 4 \times 3 a^{2} \times 4 b^{2} {= 16a}^{2} b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 8 a b + \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b + 4 a b}{6 a^{2}} = \frac{- 4 a b}{6 a^{2}} = \frac{- 2 b}{3 a} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 8 a b - \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b - 4 a b}{6 a^{2}} = \frac{- 12 a b}{6 a^{2}} = \frac{- 2 b}{a} Thus, the roots of the equation are \frac{- 2 b}{3 a} and \frac{- 2 b}{a} .$

Answer:

The given equation is $a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = $a^{2} b^{2}$ , B = $- (4 b^{4} - 3 a^{4})$ and C = $- 12 a^{2} b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (4 b^{4} - 3 a^{4})]}^{2} - 4 \times a^{2} b^{2} \times (- 12 a^{2} b^{2}) = 16 b^{8} - 24 a^{4} b^{4} + 9 a^{8} + 48 a^{4} b^{4} = 16 b^{8} + 24 a^{4} b^{4} + 9 a^{8} = {(4 b^{4} + 3 a^{4})}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(4 b^{4} + 3 a^{4})}^{2}} = 4 b^{4} + 3 a^{4}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (4 b^{4} - 3 a^{4})] + (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{8 b^{4}}{2 a^{2} b^{2}} = \frac{4 b^{2}}{a^{2}} β = \frac{- B - \sqrt{D}}{2 A} = = \frac{- [- (4 b^{4} - 3 a^{4})] - (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{- 6 a^{4}}{2 a^{2} b^{2}} = - \frac{3 a^{2}}{b^{2}}$

Hence, $\frac{4 b^{2}}{a^{2}}$ and $- \frac{3 a^{2}}{b^{2}}$ are the roots of the given equation.

Question 36:

The given equation is $a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = $a^{2} b^{2}$ , B = $- (4 b^{4} - 3 a^{4})$ and C = $- 12 a^{2} b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (4 b^{4} - 3 a^{4})]}^{2} - 4 \times a^{2} b^{2} \times (- 12 a^{2} b^{2}) = 16 b^{8} - 24 a^{4} b^{4} + 9 a^{8} + 48 a^{4} b^{4} = 16 b^{8} + 24 a^{4} b^{4} + 9 a^{8} = {(4 b^{4} + 3 a^{4})}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(4 b^{4} + 3 a^{4})}^{2}} = 4 b^{4} + 3 a^{4}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (4 b^{4} - 3 a^{4})] + (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{8 b^{4}}{2 a^{2} b^{2}} = \frac{4 b^{2}}{a^{2}} β = \frac{- B - \sqrt{D}}{2 A} = = \frac{- [- (4 b^{4} - 3 a^{4})] - (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{- 6 a^{4}}{2 a^{2} b^{2}} = - \frac{3 a^{2}}{b^{2}}$

Hence, $\frac{4 b^{2}}{a^{2}}$ and $- \frac{3 a^{2}}{b^{2}}$ are the roots of the given equation.

Answer:

$Given: 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 12 a b, B = - (9 a^{2} - 8 b^{2}) and C = - 6 a b Discriminant D is given by : D = B^{2} - 4 A C = {[- (9 a^{2} - 8 b^{2})]}^{2} - 4 \times 12 a b \times (- 6 a b) = 81 a^{4} - 144 a^{2} b^{2} + 64 b^{4} + 288 a^{2} b^{2} = 81 a^{^{4}} + 144 a^{2} b^{2} + 64 b^{4} = {(9 a^{2} + 8 b^{2})}^{2} > 0 Hence, the roots of the equation are equal. Roots α and β are given by: α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] + \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} + 9 a^{2} + 8 b^{2}}{24 a b} = \frac{18 a^{2}}{24 a b} = \frac{3 a}{4 b} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] - \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} - 9 a^{2} - 8 b^{2}}{24 a b} = \frac{- 16 b^{2}}{24 a b} = \frac{- 2 b}{3 a} Thus, the roots of the equation are \frac{3 a}{4 b} and \frac{- 2 b}{3 a} .$

Question 1:

$Given: 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 12 a b, B = - (9 a^{2} - 8 b^{2}) and C = - 6 a b Discriminant D is given by : D = B^{2} - 4 A C = {[- (9 a^{2} - 8 b^{2})]}^{2} - 4 \times 12 a b \times (- 6 a b) = 81 a^{4} - 144 a^{2} b^{2} + 64 b^{4} + 288 a^{2} b^{2} = 81 a^{^{4}} + 144 a^{2} b^{2} + 64 b^{4} = {(9 a^{2} + 8 b^{2})}^{2} > 0 Hence, the roots of the equation are equal. Roots α and β are given by: α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] + \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} + 9 a^{2} + 8 b^{2}}{24 a b} = \frac{18 a^{2}}{24 a b} = \frac{3 a}{4 b} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] - \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} - 9 a^{2} - 8 b^{2}}{24 a b} = \frac{- 16 b^{2}}{24 a b} = \frac{- 2 b}{3 a} Thus, the roots of the equation are \frac{3 a}{4 b} and \frac{- 2 b}{3 a} .$

Answer:

(i) The given equation is $2 x^{2} - 8 x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2, b = −8 and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 8)}^{2} - 4 \times 2 \times 5 = 64 - 40 = 24 > 0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = $- 2 \sqrt{6}$ and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5 x^{2} - 4 x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −4 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 5 \times 1 = 16 - 20 = - 4 < 0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5 x (x - 2) + 6 = 0 \Rightarrow 5 x^{2} - 10 x + 6 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −10 and c = 6.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 10)}^{2} - 4 \times 5 \times 6 = 100 - 120 = - 20 < 0$

Hence, the given equation has no real roots.

(v) The given equation is $12 x^{2} - 4 \sqrt{15} x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 12, b = $- 4 \sqrt{15}$ and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4 \sqrt{15})}^{2} - 4 \times 12 \times 5 = 240 - 240 = 0$

Hence, the given equation has real and equal roots.

(vi) The given equation is $x^{2} - x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −1 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 1)}^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots.

Question 2:

(i) The given equation is $2 x^{2} - 8 x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2, b = −8 and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 8)}^{2} - 4 \times 2 \times 5 = 64 - 40 = 24 > 0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = $- 2 \sqrt{6}$ and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5 x^{2} - 4 x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −4 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 5 \times 1 = 16 - 20 = - 4 < 0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5 x (x - 2) + 6 = 0 \Rightarrow 5 x^{2} - 10 x + 6 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −10 and c = 6.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 10)}^{2} - 4 \times 5 \times 6 = 100 - 120 = - 20 < 0$

Hence, the given equation has no real roots.

(v) The given equation is $12 x^{2} - 4 \sqrt{15} x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 12, b = $- 4 \sqrt{15}$ and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4 \sqrt{15})}^{2} - 4 \times 12 \times 5 = 240 - 240 = 0$

Hence, the given equation has real and equal roots.

(vi) The given equation is $x^{2} - x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −1 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 1)}^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots.

Answer:

The given equation is $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ .
$∴ D = {[2 (a + b)]}^{2} - 4 \times 2 (a^{2} + b^{2}) \times 1 = 4 (a^{2} + 2 a b + b^{2}) - 8 (a^{2} + b^{2}) = 4 a^{2} + 8 a b + 4 b^{2} - 8 a^{2} - 8 b^{2} = - 4 a^{2} + 8 a b - 4 b^{2} = - 4 (a^{2} - 2 a b + b^{2}) = - 4 {(a - b)}^{2} < 0$
Hence, the given equation has no real roots.

Question 3:

The given equation is $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ .
$∴ D = {[2 (a + b)]}^{2} - 4 \times 2 (a^{2} + b^{2}) \times 1 = 4 (a^{2} + 2 a b + b^{2}) - 8 (a^{2} + b^{2}) = 4 a^{2} + 8 a b + 4 b^{2} - 8 a^{2} - 8 b^{2} = - 4 a^{2} + 8 a b - 4 b^{2} = - 4 (a^{2} - 2 a b + b^{2}) = - 4 {(a - b)}^{2} < 0$
Hence, the given equation has no real roots.

Answer:

$Given: x^{2} + p x - q^{2} = 0 Here, a = 1, b = p and c = - q^{2} Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 1 \times (- q^{2}) = (p^{2} + 4 q^{2}) > 0 D > 0 for all real values of p and q . Thus, the roots of the equation are real .$

Question 4:

$Given: x^{2} + p x - q^{2} = 0 Here, a = 1, b = p and c = - q^{2} Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 1 \times (- q^{2}) = (p^{2} + 4 q^{2}) > 0 D > 0 for all real values of p and q . Thus, the roots of the equation are real .$

Answer:

$Given: 3 x^{2} + 2 k x + 27 = 0 Here, a = 3, b = 2 k and c = 27 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 3 \times 27 = 0 \Rightarrow 4 k^{2} - 324 = 0 \Rightarrow 4 k^{2} = 324 \Rightarrow k^{2} = 81 \Rightarrow k = \pm 9 ∴ k = 9 or k = - 9$

Question 5:

$Given: 3 x^{2} + 2 k x + 27 = 0 Here, a = 3, b = 2 k and c = 27 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 3 \times 27 = 0 \Rightarrow 4 k^{2} - 324 = 0 \Rightarrow 4 k^{2} = 324 \Rightarrow k^{2} = 81 \Rightarrow k = \pm 9 ∴ k = 9 or k = - 9$

Answer:

The given equation is

$k x (x - 2 \sqrt{5}) + 10 = 0 \Rightarrow k x^{2} - 2 \sqrt{5} k x + 10 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = k, b = $- 2 \sqrt{5} k$ and c = 10.

$∴ D = b^{2} - 4 a c = {(- 2 \sqrt{5} k)}^{2} - 4 \times k \times 10 = 20 k^{2} - 40 k$

The given equation will have real and equal roots if D = 0.

$∴ 20 k^{2} - 40 k = 0 \Rightarrow 20 k (k - 2) = 0 \Rightarrow k = 0 or k - 2 = 0 \Rightarrow k = 0 or k = 2$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

Question 6:

The given equation is

$k x (x - 2 \sqrt{5}) + 10 = 0 \Rightarrow k x^{2} - 2 \sqrt{5} k x + 10 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = k, b = $- 2 \sqrt{5} k$ and c = 10.

$∴ D = b^{2} - 4 a c = {(- 2 \sqrt{5} k)}^{2} - 4 \times k \times 10 = 20 k^{2} - 40 k$

The given equation will have real and equal roots if D = 0.

$∴ 20 k^{2} - 40 k = 0 \Rightarrow 20 k (k - 2) = 0 \Rightarrow k = 0 or k - 2 = 0 \Rightarrow k = 0 or k = 2$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

Answer:

The given equation is $4 x^{2} + p x + 3 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 4, b = p and c = 3.

$∴ D = b^{2} - 4 a c = p^{2} - 4 \times 4 \times 3 = p^{2} - 48$

The given equation will have real and equal roots if D = 0.

$∴ p^{2} - 48 = 0 \Rightarrow p^{2} = 48 \Rightarrow p = \pm \sqrt{48} = \pm 4 \sqrt{3}$

Hence, $4 \sqrt{3}$ and $- 4 \sqrt{3}$ are the required values of p.

Question 7:

The given equation is $4 x^{2} + p x + 3 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 4, b = p and c = 3.

$∴ D = b^{2} - 4 a c = p^{2} - 4 \times 4 \times 3 = p^{2} - 48$

The given equation will have real and equal roots if D = 0.

$∴ p^{2} - 48 = 0 \Rightarrow p^{2} = 48 \Rightarrow p = \pm \sqrt{48} = \pm 4 \sqrt{3}$

Hence, $4 \sqrt{3}$ and $- 4 \sqrt{3}$ are the required values of p.

Answer:

The given equation is $9 x^{2} - 3 k x + k = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 9, b = −3k and c = k.

$∴ D = b^{2} - 4 a c = {(- 3 k)}^{2} - 4 \times 9 \times k = 9 k^{2} - 36 k$

The given equation will have real and equal roots if D = 0.

$∴ 9 k^{2} - 36 k = 0 \Rightarrow 9 k (k - 4) = 0 \Rightarrow k = 0 or k - 4 = 0 \Rightarrow k = 0 or k = 4$

But, k ≠ 0 (Given)

Hence, the required value of k is 4.

Question 8:

The given equation is $9 x^{2} - 3 k x + k = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 9, b = −3k and c = k.

$∴ D = b^{2} - 4 a c = {(- 3 k)}^{2} - 4 \times 9 \times k = 9 k^{2} - 36 k$

The given equation will have real and equal roots if D = 0.

$∴ 9 k^{2} - 36 k = 0 \Rightarrow 9 k (k - 4) = 0 \Rightarrow k = 0 or k - 4 = 0 \Rightarrow k = 0 or k = 4$

But, k ≠ 0 (Given)

Hence, the required value of k is 4.

Answer:

(i)
The given equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3k +1, b = 2(k + 1) and c = 1.

$∴ D = b^{2} - 4 a c = {[2 (k + 1)]}^{2} - 4 \times (3 k + 1) \times 1 = 4 (k^{2} + 2 k + 1) - 4 (3 k + 1) = 4 k^{2} + 8 k + 4 - 12 k - 4$
$= 4 k^{2} - 4 k$

The given equation will have real and equal roots if D = 0.

$∴ 4 k^{2} - 4 k = 0 \Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Hence, 0 and 1 are the required values of k.

(ii)
$x^{2} + k (2 x + k - 1) + 2 = 0 x^{2} + 2 k x + k^{2} - k + 2 = 0 For real and equal roots D = 0 D = b^{2} - 4 a c = 0 D = {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) D = 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k - 8 = 0 \Rightarrow k = 2$

Question 9:

(i)
The given equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3k +1, b = 2(k + 1) and c = 1.

$∴ D = b^{2} - 4 a c = {[2 (k + 1)]}^{2} - 4 \times (3 k + 1) \times 1 = 4 (k^{2} + 2 k + 1) - 4 (3 k + 1) = 4 k^{2} + 8 k + 4 - 12 k - 4$
$= 4 k^{2} - 4 k$

The given equation will have real and equal roots if D = 0.

$∴ 4 k^{2} - 4 k = 0 \Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Hence, 0 and 1 are the required values of k.

(ii)
$x^{2} + k (2 x + k - 1) + 2 = 0 x^{2} + 2 k x + k^{2} - k + 2 = 0 For real and equal roots D = 0 D = b^{2} - 4 a c = 0 D = {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) D = 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k - 8 = 0 \Rightarrow k = 2$

Answer:

The given equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2p +1, b = −(7p + 2) and c = 7p − 3.

$∴ D = b^{2} - 4 a c = {[- (7 p + 2)]}^{2} - 4 \times (2 p + 1) \times (7 p - 3) = (49 p^{2} + 28 p + 4) - 4 (14 p^{2} + p - 3) = 49 p^{2} + 28 p + 4 - 56 p^{2} - 4 p + 12$
$= - 7 p^{2} + 24 p + 16$

The given equation will have real and equal roots if D = 0.

$∴ - 7 p^{2} + 24 p + 16 = 0 \Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow 7 p^{2} - 28 p + 4 p - 16 = 0 \Rightarrow 7 p (p - 4) + 4 (p - 4) = 0$
$\Rightarrow (p - 4) (7 p + 4) = 0 \Rightarrow p - 4 = 0 or 7 p + 4 = 0 \Rightarrow p = 4 or p = - \frac{4}{7}$

Hence, 4 and $- \frac{4}{7}$ are the required values of p.

Question 10:

The given equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2p +1, b = −(7p + 2) and c = 7p − 3.

$∴ D = b^{2} - 4 a c = {[- (7 p + 2)]}^{2} - 4 \times (2 p + 1) \times (7 p - 3) = (49 p^{2} + 28 p + 4) - 4 (14 p^{2} + p - 3) = 49 p^{2} + 28 p + 4 - 56 p^{2} - 4 p + 12$
$= - 7 p^{2} + 24 p + 16$

The given equation will have real and equal roots if D = 0.

$∴ - 7 p^{2} + 24 p + 16 = 0 \Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow 7 p^{2} - 28 p + 4 p - 16 = 0 \Rightarrow 7 p (p - 4) + 4 (p - 4) = 0$
$\Rightarrow (p - 4) (7 p + 4) = 0 \Rightarrow p - 4 = 0 or 7 p + 4 = 0 \Rightarrow p = 4 or p = - \frac{4}{7}$

Hence, 4 and $- \frac{4}{7}$ are the required values of p.

Answer:

The given equation is $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = p +1, b = −6(p + 1) and c = 3(p + 9).

$∴ D = b^{2} - 4 a c = {[- 6 (p + 1)]}^{2} - 4 \times (p + 1) \times 3 (p + 9) = 12 (p + 1) [3 (p + 1) - (p + 9)] = 12 (p + 1) (2 p - 6)$

The given equation will have real and equal roots if D = 0.

$∴ 12 (p + 1) (2 p - 6) = 0 \Rightarrow p + 1 = 0 or 2 p - 6 = 0 \Rightarrow p = - 1 or p = 3$

But, $p \neq - 1$ (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4 x^{2} - 24 x + 36 = 0$ .

$4 x^{2} - 24 x + 36 = 0 \Rightarrow 4 (x^{2} - 6 x + 9) = 0 \Rightarrow {(x - 3)}^{2} = 0 \Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

Hence, 3 is the repeated root of this equation.

Question 11:

The given equation is $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = p +1, b = −6(p + 1) and c = 3(p + 9).

$∴ D = b^{2} - 4 a c = {[- 6 (p + 1)]}^{2} - 4 \times (p + 1) \times 3 (p + 9) = 12 (p + 1) [3 (p + 1) - (p + 9)] = 12 (p + 1) (2 p - 6)$

The given equation will have real and equal roots if D = 0.

$∴ 12 (p + 1) (2 p - 6) = 0 \Rightarrow p + 1 = 0 or 2 p - 6 = 0 \Rightarrow p = - 1 or p = 3$

But, $p \neq - 1$ (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4 x^{2} - 24 x + 36 = 0$ .

$4 x^{2} - 24 x + 36 = 0 \Rightarrow 4 (x^{2} - 6 x + 9) = 0 \Rightarrow {(x - 3)}^{2} = 0 \Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

Hence, 3 is the repeated root of this equation.

Answer:

It is given that −5 is a root of the quadratic equation $2 x^{2} + p x - 15 = 0$ .

$∴ 2 {(- 5)}^{2} + p \times (- 5) - 15 = 0 \Rightarrow - 5 p + 35 = 0 \Rightarrow p = 7$

The roots of the equation $p x^{2} + p x + k = 0$ = 0 are equal.

$∴ D = 0 \Rightarrow p^{2} - 4 p k = 0 \Rightarrow {(7)}^{2} - 4 \times 7 \times k = 0 \Rightarrow 49 - 28 k = 0$
$\Rightarrow k = \frac{49}{28} = \frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$ .

Question 12:

It is given that −5 is a root of the quadratic equation $2 x^{2} + p x - 15 = 0$ .

$∴ 2 {(- 5)}^{2} + p \times (- 5) - 15 = 0 \Rightarrow - 5 p + 35 = 0 \Rightarrow p = 7$

The roots of the equation $p x^{2} + p x + k = 0$ = 0 are equal.

$∴ D = 0 \Rightarrow p^{2} - 4 p k = 0 \Rightarrow {(7)}^{2} - 4 \times 7 \times k = 0 \Rightarrow 49 - 28 k = 0$
$\Rightarrow k = \frac{49}{28} = \frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$ .

Answer:

It is given that 3 is a root of the quadratic equation $x^{2} - x + k = 0$ .

$∴ {(3)}^{2} - 3 + k = 0 \Rightarrow k + 6 = 0 \Rightarrow k = - 6$

The roots of the equation $x^{2} + 2 k x + (k^{2} + 2 k + p) = 0$ are equal.

$∴ D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 1 \times (k^{2} + 2 k + p) = 0 \Rightarrow 4 k^{2} - 4 k^{2} - 8 k - 4 p = 0 \Rightarrow - 8 k - 4 p = 0$
$\Rightarrow p = \frac{8 k}{- 4} = - 2 k \Rightarrow p = - 2 \times (- 6) = 12$

Hence, the value of p is 12.

Question 13:

It is given that 3 is a root of the quadratic equation $x^{2} - x + k = 0$ .

$∴ {(3)}^{2} - 3 + k = 0 \Rightarrow k + 6 = 0 \Rightarrow k = - 6$

The roots of the equation $x^{2} + 2 k x + (k^{2} + 2 k + p) = 0$ are equal.

$∴ D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 1 \times (k^{2} + 2 k + p) = 0 \Rightarrow 4 k^{2} - 4 k^{2} - 8 k - 4 p = 0 \Rightarrow - 8 k - 4 p = 0$
$\Rightarrow p = \frac{8 k}{- 4} = - 2 k \Rightarrow p = - 2 \times (- 6) = 12$

Hence, the value of p is 12.

Answer:

It is given that −4 is a root of the quadratic equation $x^{2} + 2 x + 4 p = 0$ .

$∴ {(- 4)}^{2} + 2 \times (- 4) + 4 p = 0 \Rightarrow 16 - 8 + 4 p = 0 \Rightarrow 4 p + 8 = 0 \Rightarrow p = - 2$

The equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots.

$∴ D = 0 \Rightarrow {[p (1 + 3 k)]}^{2} - 4 \times 1 \times 7 (3 + 2 k) = 0 \Rightarrow {[- 2 (1 + 3 k)]}^{2} - 28 (3 + 2 k) = 0 \Rightarrow 4 (1 + 6 k + 9 k^{2}) - 28 (3 + 2 k) = 0$
$\Rightarrow 4 (1 + 6 k + 9 k^{2} - 21 - 14 k) = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0$
$\Rightarrow (k - 2) (9 k + 10) = 0 \Rightarrow k - 2 = 0 or 9 k + 10 = 0 \Rightarrow k = 2 or k = - \frac{10}{9}$

Hence, the required value of k is 2 or $- \frac{10}{9}$ .

Question 14:

It is given that −4 is a root of the quadratic equation $x^{2} + 2 x + 4 p = 0$ .

$∴ {(- 4)}^{2} + 2 \times (- 4) + 4 p = 0 \Rightarrow 16 - 8 + 4 p = 0 \Rightarrow 4 p + 8 = 0 \Rightarrow p = - 2$

The equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots.

$∴ D = 0 \Rightarrow {[p (1 + 3 k)]}^{2} - 4 \times 1 \times 7 (3 + 2 k) = 0 \Rightarrow {[- 2 (1 + 3 k)]}^{2} - 28 (3 + 2 k) = 0 \Rightarrow 4 (1 + 6 k + 9 k^{2}) - 28 (3 + 2 k) = 0$
$\Rightarrow 4 (1 + 6 k + 9 k^{2} - 21 - 14 k) = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0$
$\Rightarrow (k - 2) (9 k + 10) = 0 \Rightarrow k - 2 = 0 or 9 k + 10 = 0 \Rightarrow k = 2 or k = - \frac{10}{9}$

Hence, the required value of k is 2 or $- \frac{10}{9}$ .

Answer:

$Given: (1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0 Here, a = (1 + m^{2}), b = 2 m c and c = (c^{2} - a^{2}) It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(2 m c)}^{2} - 4 \times (1 + m^{2}) \times (c^{2} - a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 (c^{2} - a^{2} + m^{2} c^{2} - m^{2} a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 c^{2} + 4 a^{2} - 4 m^{2} c^{2} + 4 m^{2} a^{2} = 0 \Rightarrow - 4 c^{2} + 4 a^{2} + 4 m^{2} a^{2} = 0 \Rightarrow a^{2} + m^{2} a^{2} = c^{2} \Rightarrow a^{2} (1 + m^{2}) = c^{2} \Rightarrow c^{2} = a^{2} (1 + m^{2}) Hence proved .$

Question 15:

$Given: (1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0 Here, a = (1 + m^{2}), b = 2 m c and c = (c^{2} - a^{2}) It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(2 m c)}^{2} - 4 \times (1 + m^{2}) \times (c^{2} - a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 (c^{2} - a^{2} + m^{2} c^{2} - m^{2} a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 c^{2} + 4 a^{2} - 4 m^{2} c^{2} + 4 m^{2} a^{2} = 0 \Rightarrow - 4 c^{2} + 4 a^{2} + 4 m^{2} a^{2} = 0 \Rightarrow a^{2} + m^{2} a^{2} = c^{2} \Rightarrow a^{2} (1 + m^{2}) = c^{2} \Rightarrow c^{2} = a^{2} (1 + m^{2}) Hence proved .$

Answer:

$Given: (c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0 Here, a = (c^{2} - a b), b = - 2 (a^{2} - b c), c = (b^{2} - a c) It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {- 2 (a^{2} - b c)}^{2} - 4 \times (c^{2} - a b) \times (b^{2} - a c) = 0 \Rightarrow 4 (a^{4} - 2 a^{2} b c + b^{2} c^{2}) - 4 (b^{2} c^{2} - a c^{3} - a b^{3} + a^{2} b c) = 0 \Rightarrow a^{4} - 2 a^{2} b c + b^{2} c^{2} - b^{2} c^{2} + a c^{3} + a b^{3} - a^{2} b c = 0 \Rightarrow a^{4} - 3 a^{2} b c + a c^{3} + a b^{3} = 0 \Rightarrow a (a^{3} - 3 a b c + c^{3} + b^{3}) = 0 Now, a = 0 or a^{3} - 3 a b c + c^{3} + b^{3} = 0 a = 0 {or a}^{3} + b^{3} + c^{3} = 3 a b c$

Question 16:

$Given: (c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0 Here, a = (c^{2} - a b), b = - 2 (a^{2} - b c), c = (b^{2} - a c) It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {- 2 (a^{2} - b c)}^{2} - 4 \times (c^{2} - a b) \times (b^{2} - a c) = 0 \Rightarrow 4 (a^{4} - 2 a^{2} b c + b^{2} c^{2}) - 4 (b^{2} c^{2} - a c^{3} - a b^{3} + a^{2} b c) = 0 \Rightarrow a^{4} - 2 a^{2} b c + b^{2} c^{2} - b^{2} c^{2} + a c^{3} + a b^{3} - a^{2} b c = 0 \Rightarrow a^{4} - 3 a^{2} b c + a c^{3} + a b^{3} = 0 \Rightarrow a (a^{3} - 3 a b c + c^{3} + b^{3}) = 0 Now, a = 0 or a^{3} - 3 a b c + c^{3} + b^{3} = 0 a = 0 {or a}^{3} + b^{3} + c^{3} = 3 a b c$

Answer:

$Given: 2 x^{2} + p x + 8 = 0 Here, a = 2, b = p and c = 8 Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 2 \times 8 = (p^{2} - 64) If D \geq 0, t he roots of the equation will be real. \Rightarrow (p^{2} - 64) \geq 0 \Rightarrow (p + 8) (p - 8) \geq 0 \Rightarrow p \geq 8 and p \leq - 8 T hus, the roots of the equation are real for p \geq 8 and p \leq - 8 .$

Question 17:

$Given: 2 x^{2} + p x + 8 = 0 Here, a = 2, b = p and c = 8 Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 2 \times 8 = (p^{2} - 64) If D \geq 0, t he roots of the equation will be real. \Rightarrow (p^{2} - 64) \geq 0 \Rightarrow (p + 8) (p - 8) \geq 0 \Rightarrow p \geq 8 and p \leq - 8 T hus, the roots of the equation are real for p \geq 8 and p \leq - 8 .$

Answer:

$Given: (α - 12) x^{2} + 2 (α - 12) x + 2 = 0 Here, a = (α - 12), b = 2 (α - 12) and c = 2 It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {2 (α - 12)}^{2} - 4 \times (α - 12) \times 2 = 0 \Rightarrow 4 (α^{2} - 24 α + 144) - 8 (α - 12) = 0 \Rightarrow 4 α^{2} - 96 α + 576 - 8 α + 96 = 0 \Rightarrow 4 α^{2} - 104 α + 672 = 0 \Rightarrow α^{2} - 26 α + 168 = 0 \Rightarrow α^{2} - 14 α - 12 α + 168 = 0 \Rightarrow α (α - 14) - 12 (α - 14) = 0 \Rightarrow (α - 14) (α - 12) = 0 ∴ α = 14 or α = 12 If the value of α is 12, the given equation becomes non-quadratic. Therefore, the valueof α will be 14 for the equation to have equal roots.$

Question 18:

$Given: (α - 12) x^{2} + 2 (α - 12) x + 2 = 0 Here, a = (α - 12), b = 2 (α - 12) and c = 2 It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {2 (α - 12)}^{2} - 4 \times (α - 12) \times 2 = 0 \Rightarrow 4 (α^{2} - 24 α + 144) - 8 (α - 12) = 0 \Rightarrow 4 α^{2} - 96 α + 576 - 8 α + 96 = 0 \Rightarrow 4 α^{2} - 104 α + 672 = 0 \Rightarrow α^{2} - 26 α + 168 = 0 \Rightarrow α^{2} - 14 α - 12 α + 168 = 0 \Rightarrow α (α - 14) - 12 (α - 14) = 0 \Rightarrow (α - 14) (α - 12) = 0 ∴ α = 14 or α = 12 If the value of α is 12, the given equation becomes non-quadratic. Therefore, the valueof α will be 14 for the equation to have equal roots.$

Answer:

$Given: 9 x^{2} + 8 k x + 16 = 0 Here, a = 9, b = 8 k and c = 16 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(8 k)}^{2} - 4 \times 9 \times 16 = 0 \Rightarrow 64 k^{2} - 576 = 0 \Rightarrow 64 k^{2} = 576 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3 ∴ k = 3 or k = - 3$

Question 19:

$Given: 9 x^{2} + 8 k x + 16 = 0 Here, a = 9, b = 8 k and c = 16 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(8 k)}^{2} - 4 \times 9 \times 16 = 0 \Rightarrow 64 k^{2} - 576 = 0 \Rightarrow 64 k^{2} = 576 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3 ∴ k = 3 or k = - 3$

Answer:

(i) The given equation is $k x^{2} + 6 x + 1 = 0$ .

$∴ D = 6^{2} - 4 \times k \times 1 = 36 - 4 k$

The given equation has real and distinct roots if D > 0.

$∴ 36 - 4 k > 0 \Rightarrow 4 k < 36 \Rightarrow k < 9$

(ii) The given equation is $x^{2} - k x + 9 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 1 \times 9 = k^{2} - 36$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 36 > 0 \Rightarrow (k - 6) (k + 6) > 0 \Rightarrow k < - 6 or k > 6$

(iii) The given equation is $9 x^{2} + 3 k x + 4 = 0$ .

$∴ D = {(3 k)}^{2} - 4 \times 9 \times 4 = 9 k^{2} - 144$

The given equation has real and distinct roots if D > 0.

$∴ 9 k^{2} - 144 > 0 \Rightarrow 9 (k^{2} - 16) > 0 \Rightarrow (k - 4) (k + 4) > 0 \Rightarrow k < - 4 or k > 4$

(iv) The given equation is $5 x^{2} - k x + 1 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 5 \times 1 = k^{2} - 20$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 20 > 0 \Rightarrow k^{2} - {(2 \sqrt{5})}^{2} > 0 \Rightarrow (k - 2 \sqrt{5}) (k + 2 \sqrt{5}) > 0 \Rightarrow k < - 2 \sqrt{5} or k > 2 \sqrt{5}$

Question 20:

(i) The given equation is $k x^{2} + 6 x + 1 = 0$ .

$∴ D = 6^{2} - 4 \times k \times 1 = 36 - 4 k$

The given equation has real and distinct roots if D > 0.

$∴ 36 - 4 k > 0 \Rightarrow 4 k < 36 \Rightarrow k < 9$

(ii) The given equation is $x^{2} - k x + 9 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 1 \times 9 = k^{2} - 36$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 36 > 0 \Rightarrow (k - 6) (k + 6) > 0 \Rightarrow k < - 6 or k > 6$

(iii) The given equation is $9 x^{2} + 3 k x + 4 = 0$ .

$∴ D = {(3 k)}^{2} - 4 \times 9 \times 4 = 9 k^{2} - 144$

The given equation has real and distinct roots if D > 0.

$∴ 9 k^{2} - 144 > 0 \Rightarrow 9 (k^{2} - 16) > 0 \Rightarrow (k - 4) (k + 4) > 0 \Rightarrow k < - 4 or k > 4$

(iv) The given equation is $5 x^{2} - k x + 1 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 5 \times 1 = k^{2} - 20$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 20 > 0 \Rightarrow k^{2} - {(2 \sqrt{5})}^{2} > 0 \Rightarrow (k - 2 \sqrt{5}) (k + 2 \sqrt{5}) > 0 \Rightarrow k < - 2 \sqrt{5} or k > 2 \sqrt{5}$

Answer:

The given equation is $(a - b) x^{2} + 5 (a + b) x - 2 (a - b) = 0$ .

$∴ D = {[5 (a + b)]}^{2} - 4 \times (a - b) \times [- 2 (a - b)] = 25 {(a + b)}^{2} + 8 {(a - b)}^{2}$
Since a and b are real and a ≠ b, so ${(a - b)}^{2} > 0$ and ${(a + b)}^{2} > 0$ .

∴ $8 {(a - b)}^{2} > 0$ .....(1)               (Product of two positive numbers is always positive)

Also, $25 {(a + b)}^{2} > 0$              .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25 {(a + b)}^{2} + 8 {(a - b)}^{2} > 0$                  (Sum of two positive numbers is always positive)

$\Rightarrow D > 0$

Hence, the roots of the given equation are real and unequal.

Question 21:

The given equation is $(a - b) x^{2} + 5 (a + b) x - 2 (a - b) = 0$ .

$∴ D = {[5 (a + b)]}^{2} - 4 \times (a - b) \times [- 2 (a - b)] = 25 {(a + b)}^{2} + 8 {(a - b)}^{2}$
Since a and b are real and a ≠ b, so ${(a - b)}^{2} > 0$ and ${(a + b)}^{2} > 0$ .

∴ $8 {(a - b)}^{2} > 0$ .....(1)               (Product of two positive numbers is always positive)

Also, $25 {(a + b)}^{2} > 0$              .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25 {(a + b)}^{2} + 8 {(a - b)}^{2} > 0$                  (Sum of two positive numbers is always positive)

$\Rightarrow D > 0$

Hence, the roots of the given equation are real and unequal.

Answer:

Ans

Question 22:

Ans

Answer:

Ans

Question 23:

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Answer:

It is given that the roots of the equation $a x^{2} + 2 b x + c = 0$ are real.
$∴ D_{1} = {(2 b)}^{2} - 4 \times a \times c \geq 0 \Rightarrow 4 (b^{2} - a c) \geq 0 \Rightarrow b^{2} - a c \geq 0 . . . . . (1)$

Also, the roots of the equation $b x^{2} - 2 \sqrt{a c} x + b = 0$ are real.
$∴ D_{2} = {(- 2 \sqrt{a c})}^{2} - 4 \times b \times b \geq 0 \Rightarrow 4 (a c - b^{2}) \geq 0 \Rightarrow - 4 (b^{2} - a c) \geq 0$
$\Rightarrow b^{2} - a c \leq 0 . . . . . (2)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
$b^{2} - a c = 0 \Rightarrow b^{2} = a c$

Question 24:

It is given that the roots of the equation $a x^{2} + 2 b x + c = 0$ are real.
$∴ D_{1} = {(2 b)}^{2} - 4 \times a \times c \geq 0 \Rightarrow 4 (b^{2} - a c) \geq 0 \Rightarrow b^{2} - a c \geq 0 . . . . . (1)$

Also, the roots of the equation $b x^{2} - 2 \sqrt{a c} x + b = 0$ are real.
$∴ D_{2} = {(- 2 \sqrt{a c})}^{2} - 4 \times b \times b \geq 0 \Rightarrow 4 (a c - b^{2}) \geq 0 \Rightarrow - 4 (b^{2} - a c) \geq 0$
$\Rightarrow b^{2} - a c \leq 0 . . . . . (2)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
$b^{2} - a c = 0 \Rightarrow b^{2} = a c$

Answer:

Since, $x = \frac{- 1}{2}$ is a root of the quadratic equation 3x² + 2kx + 3 = 0,
then, it must satisfies the equation.

$3 {(- \frac{1}{2})}^{2} + 2 k (- \frac{1}{2}) + 3 = 0 \Rightarrow 3 (\frac{1}{4}) - k + 3 = 0 \Rightarrow \frac{3}{4} - k + 3 = 0 \Rightarrow \frac{3 - 4 k + 12}{4} = 0 \Rightarrow 3 - 4 k + 12 = 0 \Rightarrow 4 k = 15 \Rightarrow k = \frac{15}{4} Hence, the value of k is \frac{15}{4} .$

Question 25:

Since, $x = \frac{- 1}{2}$ is a root of the quadratic equation 3x² + 2kx + 3 = 0,
then, it must satisfies the equation.

$3 {(- \frac{1}{2})}^{2} + 2 k (- \frac{1}{2}) + 3 = 0 \Rightarrow 3 (\frac{1}{4}) - k + 3 = 0 \Rightarrow \frac{3}{4} - k + 3 = 0 \Rightarrow \frac{3 - 4 k + 12}{4} = 0 \Rightarrow 3 - 4 k + 12 = 0 \Rightarrow 4 k = 15 \Rightarrow k = \frac{15}{4} Hence, the value of k is \frac{15}{4} .$

Answer:

Since, $x = \frac{- 1}{3}$ is a root of the quadratic equation 2x² + 2x + k = 0,
then, it must satisfies the equation.

â€‹ $2 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + k = 0 \Rightarrow 2 (\frac{1}{9}) - \frac{2}{3} + k = 0 \Rightarrow \frac{2}{9} - \frac{2}{3} + k = 0 \Rightarrow \frac{2 - 6 + 9 k}{9} = 0 \Rightarrow - 4 + 9 k = 0 \Rightarrow 9 k = 4 \Rightarrow k = \frac{4}{9} Hence, the value of k is \frac{4}{9} .$

Question 26:

Since, $x = \frac{- 1}{3}$ is a root of the quadratic equation 2x² + 2x + k = 0,
then, it must satisfies the equation.

â€‹ $2 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + k = 0 \Rightarrow 2 (\frac{1}{9}) - \frac{2}{3} + k = 0 \Rightarrow \frac{2}{9} - \frac{2}{3} + k = 0 \Rightarrow \frac{2 - 6 + 9 k}{9} = 0 \Rightarrow - 4 + 9 k = 0 \Rightarrow 9 k = 4 \Rightarrow k = \frac{4}{9} Hence, the value of k is \frac{4}{9} .$

Answer:

Given: x² – 8x + 18 = 0

$x^{2} - 8 x + 18 = 0 Adding and subtracting {(\frac{1}{2} \times 8)}^{2}, we get \Rightarrow x^{2} - 8 x + 18 + 4^{2} - 4^{2} = 0 \Rightarrow x^{2} - 8 x + 16 + 18 - 16 = 0 \Rightarrow {(x - 4)}^{2} + 2 = 0 \Rightarrow {(x - 4)}^{2} = - 2 \Rightarrow (x - 4) = \pm \sqrt{- 2} But, \sqrt{- 2} is not a real number .$

Hence, the quadratic equation x² – 8x + 18 = 0 has no real solution.

Question 27:

Given: x² – 8x + 18 = 0

$x^{2} - 8 x + 18 = 0 Adding and subtracting {(\frac{1}{2} \times 8)}^{2}, we get \Rightarrow x^{2} - 8 x + 18 + 4^{2} - 4^{2} = 0 \Rightarrow x^{2} - 8 x + 16 + 18 - 16 = 0 \Rightarrow {(x - 4)}^{2} + 2 = 0 \Rightarrow {(x - 4)}^{2} = - 2 \Rightarrow (x - 4) = \pm \sqrt{- 2} But, \sqrt{- 2} is not a real number .$

Hence, the quadratic equation x² – 8x + 18 = 0 has no real solution.

Answer:

Let 4x² –12x – k = 0 be a quadratic equation.

It is given that, it has no real roots.

$\Rightarrow Discriminant < 0 \Rightarrow b^{2} - 4 a c < 0 \Rightarrow {(- 12)}^{2} - 4 (4) (- k) < 0 \Rightarrow 144 + 16 k < 0 \Rightarrow 16 k < - 144 \Rightarrow k < - 9$

Hence, the values of k must be less than –9.

Question 28:

Let 4x² –12x – k = 0 be a quadratic equation.

It is given that, it has no real roots.

$\Rightarrow Discriminant < 0 \Rightarrow b^{2} - 4 a c < 0 \Rightarrow {(- 12)}^{2} - 4 (4) (- k) < 0 \Rightarrow 144 + 16 k < 0 \Rightarrow 16 k < - 144 \Rightarrow k < - 9$

Hence, the values of k must be less than –9.

Answer:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 3x²– 10x + k = 0.

Product of roots = $\frac{k}{3}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{3} \Rightarrow 1 = \frac{k}{3} \Rightarrow k = 3$

Hence, the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other is 3.

Question 29:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 3x²– 10x + k = 0.

Product of roots = $\frac{k}{3}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{3} \Rightarrow 1 = \frac{k}{3} \Rightarrow k = 3$

Hence, the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other is 3.

Answer:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 5x² +13x + k = 0.

Product of roots = $\frac{k}{5}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

Hence, the value of k is 5.

Question 30:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 5x² +13x + k = 0.

Product of roots = $\frac{k}{5}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

Hence, the value of k is 5.

Answer:

Let 3x² + kx + 3 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(k)}^{2} - 4 (3) (3) = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$

Hence, the values of k are –6 and 6.

Question 31:

Let 3x² + kx + 3 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(k)}^{2} - 4 (3) (3) = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$

Hence, the values of k are –6 and 6.

Answer:

Let 9x² – 3ax + 1 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(- 3 a)}^{2} - 4 (9) (1) = 0 \Rightarrow 9 a^{2} = 36 \Rightarrow a^{2} = 4 \Rightarrow a = \pm 2$

Hence, the values of a are –2 and 2.

Question 32:

Let 9x² – 3ax + 1 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(- 3 a)}^{2} - 4 (9) (1) = 0 \Rightarrow 9 a^{2} = 36 \Rightarrow a^{2} = 4 \Rightarrow a = \pm 2$

Hence, the values of a are –2 and 2.

Answer:

Let x² + k(2x + k – 1) + 2 = 0 be a quadratic equation.

x² + k(2x + k – 1) + 2 = 0
x² + 2xk + k² – k + 2 = 0

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) = 0 \Rightarrow 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k = 8 \Rightarrow k = 2$

Hence, the value of k is 2.

Question 1:

Let x² + k(2x + k – 1) + 2 = 0 be a quadratic equation.

x² + k(2x + k – 1) + 2 = 0
x² + 2xk + k² – k + 2 = 0

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) = 0 \Rightarrow 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k = 8 \Rightarrow k = 2$

Hence, the value of k is 2.

Answer:

Let the required natural number be x.

According to the given condition,

$x + x^{2} = 156 \Rightarrow x^{2} + x - 156 = 0 \Rightarrow x^{2} + 13 x - 12 x - 156 = 0 \Rightarrow x (x + 13) - 12 (x + 13) = 0$
$\Rightarrow (x + 13) (x - 12) = 0 \Rightarrow x + 13 = 0 or x - 12 = 0 \Rightarrow x = - 13 or x = 12$

∴ x = 12 (x cannot be negative)

Hence, the required natural number is 12.

Question 2:

Let the required natural number be x.

According to the given condition,

$x + x^{2} = 156 \Rightarrow x^{2} + x - 156 = 0 \Rightarrow x^{2} + 13 x - 12 x - 156 = 0 \Rightarrow x (x + 13) - 12 (x + 13) = 0$
$\Rightarrow (x + 13) (x - 12) = 0 \Rightarrow x + 13 = 0 or x - 12 = 0 \Rightarrow x = - 13 or x = 12$

∴ x = 12 (x cannot be negative)

Hence, the required natural number is 12.

Answer:

Let the required natural number be x.

According to the given condition,

x + \sqrt{x} = 132

Putting

\sqrt{x} = y

or x = y², we get

y^{2} + y = 132 \Rightarrow y^{2} + y - 132 = 0 \Rightarrow y^{2} + 12 y - 11 y - 132 = 0 \Rightarrow y (y + 12) - 11 (y + 12) = 0

\Rightarrow (y + 12) (y - 11) = 0 \Rightarrow y + 12 = 0 or y - 11 = 0 \Rightarrow y = - 12 or y = 11

∴ y = 11 (y cannot be negative)

Now,

\sqrt{x} = 11 \Rightarrow x = {(11)}^{2} = 121

Hence, the required natural number is 121.

Question 3:

Let the required natural number be x.

According to the given condition,

x + \sqrt{x} = 132

Putting

\sqrt{x} = y

or x = y², we get

y^{2} + y = 132 \Rightarrow y^{2} + y - 132 = 0 \Rightarrow y^{2} + 12 y - 11 y - 132 = 0 \Rightarrow y (y + 12) - 11 (y + 12) = 0

\Rightarrow (y + 12) (y - 11) = 0 \Rightarrow y + 12 = 0 or y - 11 = 0 \Rightarrow y = - 12 or y = 11

∴ y = 11 (y cannot be negative)

Now,

\sqrt{x} = 11 \Rightarrow x = {(11)}^{2} = 121

Hence, the required natural number is 121.

Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

$x (28 - x) = 192 \Rightarrow 28 x - x^{2} = 192 \Rightarrow x^{2} - 28 x + 192 = 0 \Rightarrow x^{2} - 16 x - 12 x + 192 = 0$
$\Rightarrow x (x - 16) - 12 (x - 16) = 0 \Rightarrow (x - 12) (x - 16) = 0 \Rightarrow x - 12 = 0 or x - 16 = 0 \Rightarrow x = 12 or x = 16$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Question 4:

Let the required numbers be x and (28 − x).

According to the given condition,

$x (28 - x) = 192 \Rightarrow 28 x - x^{2} = 192 \Rightarrow x^{2} - 28 x + 192 = 0 \Rightarrow x^{2} - 16 x - 12 x + 192 = 0$
$\Rightarrow x (x - 16) - 12 (x - 16) = 0 \Rightarrow (x - 12) (x - 16) = 0 \Rightarrow x - 12 = 0 or x - 16 = 0 \Rightarrow x = 12 or x = 16$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
$x^{2} + {(x + 1)}^{2} = 365 \Rightarrow x^{2} + x^{2} + 2 x + 1 = 365 \Rightarrow 2 x^{2} + 2 x - 364 = 0 \Rightarrow x^{2} + x - 182 = 0$
$\Rightarrow x^{2} + 14 x - 13 x - 182 = 0 \Rightarrow x (x + 14) - 13 (x + 14) = 0 \Rightarrow (x + 14) (x - 13) = 0 \Rightarrow x + 14 = 0 or x - 13 = 0$
$\Rightarrow x = - 14 or x = 13$
∴ x = 13 (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

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