RS Aggarwal 2015 Solutions for Class 10 Math Chapter 10 - Quadratic Equations

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Page No 423:

Question 1:

Which of the following are quadratic equation in x ?
(i) $x^{2} - x + 3 = 0$
(ii) $2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0$
(iii) $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$
(iv) $\frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0$
(v) $x^{2} - 3 x - \sqrt{x} + 4 = 0$
(vi) $x - \frac{6}{x} = 3$
(vii) $x + \frac{2}{x} = x^{2}$
(viii) $x^{2} - \frac{1}{x^{2}} = 5$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$

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.

Page No 424:

Question 2:

Which of the following are the roots of $3 x^{2} + 2 x - 1 = 0$ ?
(i) −1
(ii) $\frac{1}{3}$
(iii) $- \frac{1}{2}$

Answer:

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

Page No 424:

Question 3:

Find the value of k for which x = 1 is root of the equation $x^{2} + k x + 3 = 0$ .

Answer:

$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$

Page No 424:

Question 4:

Find the values of a and b for which $x = \frac{3}{4} and x = - 2$ are the roots of the equation $a x^{2} + b x - 6 = 0 .$

Answer:

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

Page No 424:

Question 5:

$(3 x - 5) (2 x + 3) = 0$

Answer:

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

Page No 424:

Question 6:

$5 x^{2} + 4 x = 0$

Answer:

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

Page No 424:

Question 7:

$3 x^{2} - 243 = 0$

Answer:

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

Page No 424:

Question 8:

$x^{2} + 12 x + 35 = 0$

Answer:

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

Page No 424:

Question 9:

$x^{2} = 18 x - 77$

Answer:

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

Page No 424:

Question 10:

$9 x^{2} + 6 x + 1 = 0$

Answer:

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

Page No 424:

Question 11:

$4 x^{2} - 12 x + 9 = 0$

Answer:

$Given : 4 x^{2} - 12 x + 9 = 0 \Rightarrow 4 x^{2} - (6 x + 6 x) + 9 = 0 \Rightarrow 4 x^{2} - 6 x - 6 x + 9 = 0 \Rightarrow 2 x (2 x - 3) - 3 (2 x - 3) = 0 \Rightarrow (2 x - 3) (2 x - 3) = 0 \Rightarrow 2 x - 3 = 0 or 2 x - 3 = 0 \Rightarrow x = \frac{3}{2} or x = \frac{3}{2} H ence, \frac{3}{2} is the root of the equation 4 x^{2} - 12 x + 9 = 0 .$

Page No 424:

Question 12:

$6 x^{2} + 11 x + 3 = 0$

Answer:

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

Page No 424:

Question 13:

$6 x^{2} + x - 12 = 0$

Answer:

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

Page No 424:

Question 14:

$3 x^{2} - 2 x - 1 = 0$

Answer:

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

Page No 424:

Question 15:

$6 x^{2} - x - 2 = 0$

Answer:

$Given: 6 x^{2} - x - 2 = 0 \Rightarrow 6 x^{2} - (4 x - 3 x) - 2 = 0 \Rightarrow 6 x^{2} - 4 x + 3 x - 2 = 0 \Rightarrow 2 x (3 x - 2) + 1 (3 x - 2) = 0 \Rightarrow (2 x + 1) (3 x - 2) = 0 \Rightarrow 2 x + 1 = 0 or 3 x - 2 = 0 \Rightarrow x = \frac{- 1}{2} or x = \frac{2}{3} Hence, the roots of the equation are \frac{- 1}{2} and \frac{2}{3} .$

Page No 424:

Question 16:

$48 x^{2} - 13 x - 1 = 0$

Answer:

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

Page No 424:

Question 17:

$3 x^{2} + 11 x + 10 = 0$

Answer:

$Given: 3 x^{2} + 11 x + 10 = 0 \Rightarrow 3 x^{2} + (6 x + 5 x) + 10 = 0 \Rightarrow 3 x^{2} + 6 x + 5 x + 10 = 0 \Rightarrow 3 x (x + 2) + 5 (x + 2) = 0 \Rightarrow (3 x + 5) (x + 2) = 0 \Rightarrow 3 x + 5 = 0 or x + 2 = 0 \Rightarrow x = \frac{- 5}{3} or x = - 2 Hence, the roots of the equation are \frac{- 5}{3} and - 2 .$

Page No 424:

Question 18:

$4 x^{2} - 9 x = 100$

Answer:

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

Page No 424:

Question 19:

$9 x^{2} - 22 x + 8 = 0$

Answer:

$Given: 9 x^{2} - 22 x + 8 = 0 \Rightarrow 9 x^{2} - (18 x + 4 x) + 8 = 0 \Rightarrow 9 x^{2} - 18 x - 4 x + 8 = 0 \Rightarrow 9 x (x - 2) - 4 (x - 2) = 0 \Rightarrow (x - 2) (9 x - 4) = 0 \Rightarrow x - 2 = 0 or 9 x - 4 = 0 \Rightarrow x = 2 or x = \frac{4}{9} H ence, the roots of the equation are 2 and \frac{4}{9} .$

Page No 424:

Question 20:

$15 x^{2} - 28 = x$

Answer:

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

Page No 424:

Question 21:

$4 - 11 x = 3 x^{2}$

Answer:

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

Page No 424:

Question 22:

$4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$

Answer:

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

Page No 424:

Question 23:

$\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0$

Answer:

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

Page No 424:

Question 24:

$4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$

Answer:

$Given: 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0 \Rightarrow 4 \sqrt{3} x^{2} + 8 x - 3 x - 2 \sqrt{3} = 0 \Rightarrow 4 x (\sqrt{3} x + 2) - \sqrt{3} (\sqrt{3} x + 2) = 0 \Rightarrow (4 x - \sqrt{3}) (\sqrt{3} x + 2) = 0 \Rightarrow 4 x - \sqrt{3} = 0 or \sqrt{3} x + 2 = 0 \Rightarrow x = \frac{\sqrt{3}}{4} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{- 2 \sqrt{3}}{3} H ence, the roots of the equation are \frac{\sqrt{3}}{4} and \frac{- 2 \sqrt{3}}{3} .$

Page No 424:

Question 25:

$3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0$

Answer:

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

Page No 424:

Question 26:

$\sqrt{7} y^{2} - 6 y - 13 \sqrt{7} = 0$

Answer:

$Given: \sqrt{7} y^{2} - 6 y - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} y^{2} - (13 y - 7 y) - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} y^{2} + 7 y - 13 y - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} y (y + \sqrt{7}) - 13 (y + \sqrt{7}) = 0 \Rightarrow (y + \sqrt{7}) (\sqrt{7} y - 13) = 0 \Rightarrow y + \sqrt{7} = 0 or \sqrt{7} y - 13 = 0 \Rightarrow y = - \sqrt{7} or y = \frac{13}{\sqrt{7}} = \frac{13 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{13 \sqrt{7}}{7} H ence, the roots of the equation are - \sqrt{7} and \frac{13 \sqrt{7}}{7} .$

Page No 424:

Question 27:

$4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0$

Answer:

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

Page No 424:

Question 28:

$5 x - \frac{35}{x} = 18, x \neq 0$

Answer:

$Given: 5 x - \frac{35}{x} = 18 \Rightarrow 5 x^{2} - 35 = 18 x [Multiplying both side by x] \Rightarrow 5 x^{2} - 18 x - 35 = 0 \Rightarrow 5 x^{2} - (25 x - 7 x) - 35 = 0 \Rightarrow 5 x^{2} - 25 x + 7 x - 35 = 0 \Rightarrow 5 x (x - 5) + 7 (x - 5) = 0 \Rightarrow (5 x + 7) (x - 5) = 0 \Rightarrow 5 x + 7 = 0 or x - 5 = 0 \Rightarrow x = \frac{- 7}{5} or x = 5 H ence, the roots of the equation are \frac{- 7}{5} and 5 .$

Page No 424:

Question 29:

$10 x - \frac{1}{3} = 3$

Answer:

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

Page No 424:

Question 30:

$\frac{2}{x^{2}} - \frac{5}{x} + 2 = 0$

Answer:

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

Page No 424:

Question 31:

$a b x^{2} + (b^{2} - a c) x - b c = 0$

Answer:

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

Page No 424:

Question 32:

$a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0$

Answer:

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

Page No 424:

Question 33:

$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$

Answer:

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

Page No 424:

Question 34:

$4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0$

Answer:

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

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Question 35:

$\frac{1}{(x + 4)} - \frac{1}{(x - 7)} = \frac{11}{30}, (x \neq - 4, 7)$

Answer:

$Given : \frac{1}{(x + 4)} - \frac{1}{(x - 7)} = \frac{11}{30} \Rightarrow \frac{(x - 7) - (x + 4)}{(x + 4) (x - 7)} = \frac{11}{30} \Rightarrow \frac{- 11}{(x^{2} - 3 x - 28)} = \frac{11}{30} \Rightarrow \frac{- 1}{(x^{2} - 3 x - 28)} = \frac{1}{30} [O n dividing both side by 11] \Rightarrow (x^{2} - 3 x - 28) = - 30 [On cross multiplying] \Rightarrow x^{2} - 3 x + 2 = 0 \Rightarrow x^{2} - 2 x - x + 2 = 0 \Rightarrow x (x - 2) - 1 (x - 2) = 0 \Rightarrow (x - 1) (x - 2) = 0 \Rightarrow x - 1 = 0 or x - 2 = 0 \Rightarrow x = 1 or x = 2 H ence, the roots of the equation are 1 and 2 .$

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Question 36:

$\frac{1}{(x - 3)} - \frac{1}{(x + 5)} = \frac{1}{6}, (x \neq 3, - 5)$

Answer:

$Given : \frac{1}{(x - 3)} - \frac{1}{(x + 5)} = \frac{1}{6} \Rightarrow \frac{(x + 5) - (x - 3)}{(x - 3) (x + 5)} = \frac{1}{6} \Rightarrow \frac{8}{(x^{2} - 2 x - 15)} = \frac{1}{6} \Rightarrow (x^{2} - 2 x - 15) = 48 [On cross multiplying] \Rightarrow x^{2} - 2 x - 63 = 0 \Rightarrow x^{2} - (9 x - 7 x) - 63 = 0 \Rightarrow x^{2} - 9 x + 7 x - 63 = 0 \Rightarrow x (x - 9) + 7 (x - 9) = 0 \Rightarrow (x + 7) (x - 9) = 0 \Rightarrow x + 7 = 0 or x - 9 = 0 \Rightarrow x = - 7 or x = 9 H ence, the roots of the equation are - 7 and 9 .$

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Question 37:

$\frac{(x - 3)}{(x + 3)} - \frac{(x + 3)}{(x - 3)} = 6 \frac{6}{7}, (x \neq - 3, 3)$

Answer:

$Given : \frac{(x - 3)}{(x + 3)} - \frac{(x + 3)}{(x - 3)} = 6 \frac{6}{7} \Rightarrow \frac{(x - 3) (x - 3) - (x + 3) (x + 3)}{(x + 3) (x - 3)} = \frac{48}{7} \Rightarrow \frac{(x^{2} - 6 x + 9) - (x^{2} + 6 x + 9)}{(x^{2} - 9)} = \frac{48}{7} \Rightarrow \frac{- 12 x}{(x^{2} - 9)} = \frac{48}{7} \Rightarrow \frac{- x}{(x^{2} - 9)} = \frac{4}{7} [On dividing both sides by 12] \Rightarrow (4 x^{2} - 36) = - 7 x [On cross multiplying] \Rightarrow 4 x^{2} + 7 x - 36 = 0 \Rightarrow 4 x^{2} + 16 x - 9 x - 36 = 0 \Rightarrow 4 x (x + 4) - 9 (x + 4) = 0 \Rightarrow (4 x - 9) (x + 4) = 0 \Rightarrow 4 x - 9 = 0 or x + 4 = 0 \Rightarrow x = \frac{9}{4} or x = - 4 H ence, the roots of the equation are \frac{9}{4} and - 4.$

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Question 38:

$\frac{2 x}{(x - 4)} + \frac{(2 x - 5)}{(x - 3)} = \frac{25}{3}, (x \neq 4, 3)$

Answer:

$Given : \frac{2 x}{(x - 4)} + \frac{(2 x - 5)}{(x - 3)} = \frac{25}{3} \Rightarrow \frac{2 x (x - 3) + (2 x - 5) (x - 4)}{(x - 4) (x - 3)} = \frac{25}{3} \Rightarrow \frac{2 x^{2} - 6 x + 2 x^{2} - 5 x - 8 x + 20}{(x^{2} - 7 x + 12)} = \frac{25}{3} \Rightarrow \frac{4 x^{2} - 19 x + 20}{(x^{2} - 7 x + 12)} = \frac{25}{3} \Rightarrow 12 x^{2} - 57 x + 60 = 25 x^{2} - 175 x + 300 [On cross multiplying] \Rightarrow - 13 x^{2} + 118 x - 240 = 0 \Rightarrow 13 x^{2} - (78 + 40) x + 240 = 0 \Rightarrow 13 x^{2} - 78 x - 40 x + 240 = 0 \Rightarrow 13 x (x - 6) - 40 (x - 6) = 0 \Rightarrow (13 x - 40) (x - 6) = 0 \Rightarrow 13 x - 40 = 0 or x - 6 = 0 \Rightarrow x = \frac{40}{13} or x = 6 H ence, the roots of the equation are \frac{40}{13} and 6.$

Page No 425:

Question 39:

$\frac{(x + 3)}{(x - 2)} - \frac{(1 - x)}{x} = \frac{17}{4}, (x = 0, 2)$

Answer:

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

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Question 40:

$\frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x}, (x \neq 2, 1)$

Answer:

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

Page No 425:

Question 41:

$\frac{1}{(x - 2)} + \frac{1}{x} = \frac{8}{(2 x + 5)}, x \neq 0, \frac{- 5}{2}$

Answer:

$Given : \frac{1}{(x - 2)} + \frac{1}{x} = \frac{8}{(2 x + 5)} \Rightarrow \frac{x + (x - 2)}{x (x - 2)} = \frac{8}{(2 x + 5)} \Rightarrow \frac{(2 x - 2)}{x (x - 2)} = \frac{8}{(2 x + 5)} \Rightarrow (2 x - 2) (2 x + 5) = 8 x (x - 2) [On cross multiplying] \Rightarrow 4 x^{2} + 6 x - 10 = 8 x^{2} - 16 x \Rightarrow 4 x^{2} - 22 x + 10 = 0 \Rightarrow 2 (2 x^{2} - 11 x + 5) = 0 \Rightarrow 2 x^{2} - 11 x + 5 = 0 \Rightarrow 2 x^{2} - (10 + 1) x + 5 = 0 \Rightarrow 2 x^{2} - 10 x - x + 5 = 0 \Rightarrow 2 x (x - 5) - 1 (x - 5) = 0 \Rightarrow (x - 5) (2 x - 1) = 0 \Rightarrow x - 5 = 0 or 2 x - 1 = 0 \Rightarrow x = 5 or x = \frac{1}{2} Hence, the roots of the equation are 5 and \frac{1}{2} .$

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Question 42:

${(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0, (x \neq - 1)$

Answer:

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

Page No 425:

Question 43:

$2 (\frac{x - 1}{x + 3}) - 7 (\frac{x + 3}{x - 2}) = 5, (x \neq - 3, 1)$

Answer:

$Given : 2 (\frac{x - 1}{x + 3}) - 7 (\frac{x + 3}{x - 1}) = 5 P utting \frac{x - 1}{x + 3} = y, we get : 2 y - \frac{7}{y} = 5 \Rightarrow \frac{2 y^{2} - 7}{y} = 5 \Rightarrow 2 y^{2} - 7 = 5 y \Rightarrow 2 y^{2} - 5 y - 7 = 0 \Rightarrow 2 y^{2} - (7 - 2) y - 7 = 0 \Rightarrow 2 y^{2} - 7 y + 2 y - 7 = 0 \Rightarrow y (2 y - 7) + 1 (2 y - 7) = 0 \Rightarrow (2 y - 7) (y + 1) = 0 \Rightarrow 2 y - 7 = 0 or y + 1 = 0 \Rightarrow y = \frac{7}{2} or y = - 1 Case I If y = \frac{7}{2}, we get : \frac{x - 1}{x + 3} = \frac{7}{2} \Rightarrow 2(x - 1) = 7 (x + 3) [On cross multiplying] \Rightarrow 2 x - 2 = 7 x + 21 \Rightarrow - 5 x = 23 \Rightarrow x = \frac{- 23}{5} Case II If y = - 1, we get : \frac{x - 1}{x + 3} = - 1 \Rightarrow x - 1 = - 1 (x + 3) \Rightarrow x - 1 = - x - 3 \Rightarrow 2 x = - 2 \Rightarrow x = - 1 Hence, the roots of the equation are \frac{- 23}{5} and - 1.$

Page No 425:

Question 44:

$2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5, (x \neq - 3, \frac{1}{2})$

Answer:

$Given : 2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5 P utting \frac{2 x - 1}{x + 3} = y, we get: 2 y - \frac{3}{y} = 5 \Rightarrow \frac{2 y^{2} - 3}{y} = 5 \Rightarrow 2 y^{2} - 3 = 5 y [On cross multiplying] \Rightarrow 2 y^{2} - 5 y - 3 = 0 \Rightarrow 2 y^{2} - (6 - 1) y - 3 = 0 \Rightarrow 2 y^{2} - 6 y + y - 3 = 0 \Rightarrow 2 y (y - 3) + 1 (y - 3) = 0 \Rightarrow (y - 3) (2 y + 1) = 0 \Rightarrow y - 3 = 0 or 2 y + 1 = 0 \Rightarrow y = 3 or y = - \frac{1}{2} Case I If y = 3, we get : \frac{2 x - 1}{x + 3} = 3 \Rightarrow 2 x - 1 = 3 (x + 3) [On cross multiplying] \Rightarrow 2 x - 1 = 3 x + 9 \Rightarrow - x = 10 \Rightarrow x = - 10 Case II If y = - \frac{1}{2}, we get : \Rightarrow \frac{2 x - 1}{x + 3} = - \frac{1}{2} \Rightarrow 2 (2 x - 1) = - 1 (x + 3) \Rightarrow 4 x - 2 = - x - 3 \Rightarrow 5 x = - 1 \Rightarrow x = - \frac{1}{5} Hence, the roots of the equation are - 10 and - \frac{1}{5} .$

Page No 425:

Question 45:

$(\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3, (x \neq \frac{- 1}{2}, \frac{3}{4})$

Answer:

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

Page No 425:

Question 46:

$\frac{a}{(x - b)} + \frac{b}{(x - a)} = 2, (x \neq b, a)$

Answer:

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

Page No 425:

Question 47:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b), (x \neq \frac{1}{a}, \frac{1}{b})$

Answer:

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

Page No 425:

Question 48:

$3^{(x + 2)} + 3^{- x} = 10$

Answer:

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

Page No 425:

Question 49:

$4^{(x + 1)} + 4^{(1 - x)} = 10$

Answer:

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

Page No 425:

Question 50:

$2^{2 x} - 3 . 2^{(x + 2)} + 32 = 0$

Answer:

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

Page No 431:

Question 1:

$2 x^{2} - 7 x + 6 = 0$

Answer:

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

Page No 431:

Question 2:

$3 x^{2} - 2 x + 8 = 0$

Answer:

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

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Question 3:

$2 x^{2} - 5 \sqrt{2} x + 4 = 0$

Answer:

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

Page No 431:

Question 4:

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

Answer:

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

Page No 431:

Question 5:

$1 - x = 2 x^{2}$

Answer:

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

Page No 431:

Question 6:

$x^{2} = 4 x - c$

Answer:

$Given : x^{2} = 4 x - c ⇒ x^{2} - 4 x + c = 0 Here, a = 1 b = - 4 c = c Discriminant D is given by : D = b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 1 \times c = 16 - 4 c$

Page No 431:

Question 7:

$6 x^{2} + 7 x - 10 = 0$

Answer:

$Given: 6 x^{2} + 7 x - 10 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 6, b = 7 and c = - 10 Discriminant D is given by : D = (b^{2} - 4 a c) = [7^{2} - 4 \times 6 \times (- 10)] = 49 + 240 = 289 > 0 Thus, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 7 + \sqrt{289}}{2 \times 6} = \frac{- 7 + \sqrt{289}}{12} = \frac{- 7 + 17}{12} = \frac{10}{12} = \frac{5}{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 7 - \sqrt{289}}{2 \times 6} = \frac{- 7 - \sqrt{289}}{12} = \frac{- 7 - 17}{12} = \frac{- 24}{12} = - 2 Hence, the roots of the equation are \frac{5}{6} and - 2 .$

Page No 431:

Question 8:

$2 x^{2} - 9 x + 7 = 0$

Answer:

$Given : \begin{array}{l} 2 x^{2} - 9 x + 7 = 0 \end{array} On c omparing it with a x^{2} + b x + c = 0, we get: a = 2, b = - 9 and c = 7 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 9)}^{2} - 4 \times 2 \times 7 = 81 - 56 = 25 = 25 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 9) - \sqrt{25}}{2 \times 2} = \frac{9 - 5}{4} = \frac{4}{4} = 1 β = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 9) + \sqrt{25}}{2 \times 2} = \frac{9 + 5}{4} = \frac{14}{4} = \frac{7}{2} Hence, the roots of the equation are 1 and \frac{7}{2} .$

Page No 431:

Question 9:

$2 x^{2} + x - 6 = 0$

Answer:

$Given : 2 x^{2} + x - 6 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 2, b = 1 and c = - 6 Discriminant D is given by : D = (b^{2} - 4 a c) {= 1}^{2} - 4 \times 2 \times (- 6) = 1 + 48 = 49 = 49 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 1 + \sqrt{49}}{2 \times 2} = \frac{- 1 + 7}{4} = \frac{6}{4} = \frac{3}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 1 - \sqrt{49}}{2 \times 2} = \frac{- 1 - 7}{4} = \frac{- 8}{4} = - 2 Thus, the roots of the equation are \frac{3}{2} and - 2 .$

Page No 431:

Question 10:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 4 x - 1 = 0$

Answer:

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

Page No 431:

Question 11:

$x^{2} - 6 x + 4 = 0$

Answer:

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

Page No 431:

Question 12:

$x^{2} - 7 x - 5 = 0$

Answer:

$Given : x^{2} - 7 x - 5 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 1, b = - 7 and c = - 5 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 7)}^{2} - 4 \times 1 \times (- 5) = 49 + 20 = 69 > 0 Hence,the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 7) + \sqrt{69}}{2 \times 1} = \frac{7 + \sqrt{69}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 7) - \sqrt{69}}{2} = \frac{7 - \sqrt{69}}{2} Thus, the roots of the equation are (\frac{7 + \sqrt{69}}{2}) and (\frac{7 - \sqrt{69}}{2}) .$

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Question 13:

$5 x^{2} - 19 x + 17 = 0$

Answer:

$Given : 5 x^{2} - 19 x + 17 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 5, b = - 19 and c = 17 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 19)}^{2} - 4 \times 5 \times 17 = 361 - 340 = 21 > 0 Hence, the roots of the equation are real . Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 19) + \sqrt{21}}{2 \times 5} = \frac{- (- 19) + \sqrt{21}}{10} = \frac{19 + \sqrt{21}}{10} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 19) - \sqrt{21}}{2 \times 5} = \frac{- (- 19) - \sqrt{21}}{10} = \frac{19 - \sqrt{21}}{10} Thus, the roots of the equation are \frac{19 + \sqrt{21}}{10} and \frac{19 - \sqrt{21}}{10} .$

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Question 14:

$3 x^{2} - 32 x + 12 = 0$

Answer:

$Given: 3 x^{2} - 32 x + 12 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 3, b = - 32 and c = 12 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 32)}^{2} - 4 \times 3 \times 12 = 1024 - 144 = 880 > 0 Hence, the roots of the equation are real . Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 32) + \sqrt{880}}{2 \times 3} = \frac{32 + 4 \sqrt{55}}{6} = \frac{4 (8 + \sqrt{55})}{6} = \frac{2}{3} (8 + \sqrt{55}) = \frac{(16 + 2 \sqrt{55})}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 32) - \sqrt{880}}{2 \times 3} = \frac{32 - 4 \sqrt{55}}{6} = \frac{4 (8 - \sqrt{55})}{6} = \frac{2}{3} (8 - \sqrt{55}) = \frac{(16 - 2 \sqrt{55})}{3} Thus, the roots of the equation are \frac{(16 + 2 \sqrt{55})}{3} and \frac{(16 - 2 \sqrt{55})}{3} .$

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Question 15:

$25 x^{2} + 30 x + 7 = 0$

Answer:

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

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Question 16:

$15 x^{2} - 28 = x$

Answer:

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

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Question 17:

$4 - 11 x = 3 x^{2}$

Answer:

$Given: 4 - 11 x = 3 x^{2} ⇒ 3 x^{2} + 11 x - 4 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 3, b = 11 and c = - 4 Discriminant D is given by : D = (b^{2} - 4 a c) {= 11}^{2} - 4 \times 3 \times (- 4) = 121 + 48 = 169 = 169 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 11 + \sqrt{169}}{2 \times 3} = \frac{- 11 + 13}{6} = \frac{2}{6} = \frac{1}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 11 - \sqrt{169}}{2 \times 3} = \frac{- 11 - 13}{6} = \frac{- 24}{6} = - 4 Thus, the roots of the equation are \frac{1}{3} and - 4 .$

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Question 18:

$16 x^{2} = 24 x + 1$

Answer:

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

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Question 19:

$3 x^{2} + 2 \sqrt{5} x - 5 = 0$

Answer:

$Given: 3 x^{2} + 2 \sqrt{5} x - 5 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 3, b = 2 \sqrt{5} and c = - 5 Discriminant D is given by : D = (b^{2} - 4 a c) = {(2 \sqrt{5})}^{2} - 4 \times 3 \times (- 5) = (4 \times 5) + 60 = 20 + 60 = 80 = 80 > 0 Hence, the roots of the equation are real . Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 2 \sqrt{5} + \sqrt{80}}{2 \times 3} = \frac{- 2 \sqrt{5} + 4 \sqrt{5}}{6} = \frac{\sqrt{5} (- 2 + 4)}{6} = \frac{\sqrt{5}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 2 \sqrt{5} - \sqrt{80}}{2 \times 3} = \frac{\sqrt{5} (- 2 - 4)}{6} = \frac{\sqrt{5} (- 6)}{6} = - \sqrt{5} Thus, the roots of the equation are \frac{\sqrt{5}}{3} and - \sqrt{5} .$

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Question 20:

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

Answer:

$Given: 2 x^{2} - 2 \sqrt{6} x + 3 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 2, b = - 2 \sqrt{6} and c = 3 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 2 \sqrt{6})}^{2} - 4 \times 2 \times 3 = 24 - 24 = 0 Hence, the equation has two equal roots. R oots are given by: x = \frac{- b}{2 a} and \frac{- b}{2 a} = \frac{- (- 2 \sqrt{6})}{2 \times 2} and \frac{- (- 2 \sqrt{6})}{2 \times 2} = \frac{\sqrt{6}}{2} and \frac{\sqrt{6}}{2} Thus, the roots are \frac{\sqrt{6}}{2} and \frac{\sqrt{6}}{2} .$

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Question 21:

$\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$

Answer:

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

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Question 22:

$9 x^{2} - 4 = 0$

Answer:

$Given: 9 x^{2} - 4 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 9, b = 0 and c = - 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(0)}^{2} - 4 \times 9 \times (- 4) = 144 > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 0 + \sqrt{144}}{2 \times 9} = \frac{- 0 + 12}{18} = \frac{12}{18} = \frac{2}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 0 - \sqrt{144}}{2 \times 9} = \frac{- 0 - 12}{18} = \frac{- 12}{18} = \frac{- 2}{3} Thus, the roots of the equation are \frac{2}{3} and \frac{- 2}{3} .$

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Question 23:

$3 a^{2} x^{2} + 8 a b x + 4 b^{2} = 0, a \neq 0$

Answer:

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

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Question 24:

$p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$

Answer:

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

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Question 25:

$x^{2} - 2 a x + (a^{2} - b^{2}) = 0$

Answer:

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

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Question 26:

$a b x^{2} + (b^{2} - a c) x - b c = 0$

Answer:

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

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Question 27:

$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$ , where a $\neq$ 0 and b $\neq$ 0

Answer:

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

Page No 439:

Question 1:

$x^{2} + 8 x + 16 = 0$

Answer:

$Given: x^{2} + 8 x + 16 = 0 Here, a = 1, b = 8 and c = 16 Discriminant D is given by : D = (b^{2} - 4 a c) = {(8)}^{2} - 4 \times 1 \times 16 = 64 - 64 = 0 Thus, the roots of the equation are equal.$

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Question 2:

$x^{2} - 6 x + 6 = 0$

Answer:

$Given: x^{2} - 6 x + 6 = 0 Here, a = 1, b = - 6 and c = 6 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 1 \times 6 = 36 - 24 = 12 > 0 Thus, the roots of the equation are not equal.$

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Question 3:

$9 x^{2} - 6 x + 4 = 0$

Answer:

$Given: 9 x^{2} - 12 x + 4 = 0 Here, a = 9, b = - 12 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 12)}^{2} - 4 \times 9 \times 4 = 0 Thus, the roots of the equation are equal.$

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Question 4:

$9 x_{^{2}} - 6 x + 4 = 0$

Answer:

$Given: 9 x^{2} - 6 x + 4 = 0 Here, a = 9, b = - 6 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 9 \times 4 = 36 - 144 = - 108 < 0 Thus, the roots of the equation are not equal.$

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Question 5:

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

Answer:

$Given: 3 x^{2} - 2 \sqrt{6} x + 2 = 0 Here, a = 3, b = - 2 \sqrt{6} and c = 2 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0 Thus, the roots of the equation are equal .$

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Question 6:

$12 x^{2} - 4 \sqrt{15} x + 5 = 0$

Answer:

$Given: 12 x^{2} - 4 \sqrt{15} x + 5 = 0 Here, a = 12, b = - 4 \sqrt{15} and c = 5 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4 \sqrt{15})}^{2} - 4 \times 12 \times 5 = 240 - 240 = 0 Thus, the roots of the equation are equal .$

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Question 7:

Show that the roots of the equation $x^{2} + p x - q^{2} = 0$ are real for all real value of p and q.

Answer:

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

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Question 8:

Show that the equation $x^{2} + a x - 1 = 0$ has real and distinct roots for all real values of a.

Answer:

$Given: x^{2} + a x - 1 = 0 Here, a = 1, b = a and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) \Rightarrow D = a^{2} - 4 \times 1 \times (- 1) \Rightarrow D = (a^{2} + 4) F o r a l l r e a l v a l u e s o f a, w e h a v e a^{2} > 0 o r, a^{2} + 4 > 0 ∴ D > 0 for all real values of a . Thus, the roots of the equation are real and distinct for all real values of a .$

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Question 9:

Show that the equation $x^{2} - x + 2 = 0$ has no real roots.

Answer:

$Given: x^{2} - x + 2 = 0 Here, a = 1, b = - 1 and c = 2 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 1)}^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0 D < 0; therefore, the roots of the equation are not real .$

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Question 10:

Find the values of k for which the equation $k x^{2} + 2 x + 1 = 0$ has real and distinct roots.

Answer:

$Given: k x^{2} + 2 x + 1 = 0 Here, a = k, b = 2 and c = 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(2)}^{2} - 4 \times k \times 1 = 4 - 4 k If D > 0, t he roots of the equation will be real and distinct. \Rightarrow 4 - 4 k > 0 \Rightarrow 4 (1 - k) > 0 \Rightarrow (1 - k) > 0 \Rightarrow k < 1 Thus, the equation has real and distinct roots for all real values of k < 1 .$

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Question 11:

Determine the values of p for which the quadratic equation $2 x^{2} + p x + 8 = 0$ has real roots.

Answer:

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

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Question 12:

Find the value of α for which the equation $(α - 12) x^{2} + 2 (α - 12) x + 2 = 0$ has equal roots.

Answer:

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

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Question 13:

If the equation $(1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0$ has equal roots, prove that $c^{2} = a^{2} (1 + m^{2}) .$

Answer:

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

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Question 14:

If the roots of the equation $(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0$ are real and equal, show that either $a = 0 or (a^{3} + b^{3} + c^{3}) = 3 a b c .$

Answer:

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

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Question 15:

Find the values k for which of roots of $9 x^{2} + 8 k x + 16 = 0$ are real and equal

Answer:

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

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Question 16:

Find the values of k for which the roots of the equation $(k + 4) x^{2} + (k + 1) x + 1 = 0$ are real and equal?

Answer:

$Given: (k + 4) x^{2} + (k + 1) x + 1 = 0 Here, a = (k + 4), b = (k + 1) and c = 1 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(k + 1)}^{2} - 4 \times (k + 4) \times 1 = 0 \Rightarrow k^{2} + 2 k + 1 - 4 k - 16 = 0 \Rightarrow k^{2} - 2 k - 15 = 0 \Rightarrow k^{2} - 5 k + 3 k - 15 = 0 \Rightarrow k (k - 5) + 3 (k - 5) = 0 \Rightarrow (k - 5) (k + 3) = 0 \Rightarrow k = 5 or k = - 3$

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Question 17:

For what values of k are the roots of the quadratic equation $3 x^{2} + 2 k x + 27 = 0$ real and equal?

Answer:

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

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Question 1:

The sum of two numbers is 8. Determine the numbers, if the sum of their reciprocals is $\frac{8}{15} .$

Answer:

$Let the numbers be x and (8 - x) . According to the question : \frac{1}{x} + \frac{1}{8 - x} = \frac{8}{15} \begin{array}{l} \Rightarrow \frac{8 - x + x}{x (8 - x)} = \frac{8}{15} \\ \Rightarrow \frac{8^{1}}{8 x - x^{2}} = \frac{8^{1}}{15} \Rightarrow \frac{1}{8 x - x^{2}} = \frac{1}{15} \Rightarrow 8 x - x^{2} = 15 \end{array} \Rightarrow x^{2} - 8 x + 15 = 0 \Rightarrow x^{2} - (5 + 3) x + 15 = 0 \Rightarrow x^{2} - 5 x - 3 x + 15 = 0 \Rightarrow x (x - 5) - 3 (x - 5) = 0 \Rightarrow (x - 5) (x - 3) = 0 \Rightarrow x - 5 = 0 or x - 3 = 0 \Rightarrow x = 5 or x = 3 If x = 5, the numbers are 5 and 3. {∵ (8 - 5) = 3} If x = 3, the numbers are 3 and 5. {∵ (8 - 3) = 5} Hence, the numbers are 5 and 3.$

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Question 2:

The difference of two numbers is 4. If the difference of their reciprocals is $\frac{4}{21}$ , find the number.

Answer:

$Let the numbers be x and (x + 4) . According to the question: \frac{1}{x} - \frac{1}{x + 4} = \frac{4}{21} \Rightarrow \frac{x + 4 - x}{x (x + 4)} = \frac{4}{21} \Rightarrow \frac{4^{1}}{x^{2} + 4 x} = \frac{4^{1}}{21} \Rightarrow \frac{1}{x^{2} + 4 x} = \frac{1}{21} \Rightarrow x^{2} + 4 x = 21 \Rightarrow x^{2} + 4 x - 21 = 0 \Rightarrow x^{2} + (7 - 3) x - 21 = 0 \Rightarrow x^{2} + 7 x - 3 x - 21 = 0 \Rightarrow x (x + 7) - 3 (x + 7) = 0 \Rightarrow (x + 7) (x - 3) = 0 \Rightarrow x + 7 = 0 or x - 3 = 0 \Rightarrow x = - 7 or x = 3 If x = - 7, the numbers are - 7 and -3. (∵ - 7 + 4 = - 3) If x = 3, the numbers are 3 and 7. (∵ 3 + 4 = 7) Hence, the numbers are (- 7, - 3) or (3, 7) .$

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Question 3:

The sum of two numbers is 18 and the sum of their reciprocals is $\frac{1}{4}$ . Find the numbers.

Answer:

$Let the numbers be x and (18 - x) . According to the question: \frac{1}{x} + \frac{1}{18 - x} = \frac{1}{4} \Rightarrow \frac{18 - x + x}{x (18 - x)} = \frac{1}{4} \Rightarrow \frac{18}{18 x - x^{2}} = \frac{1}{4} \Rightarrow 18 x - x^{2} = 72 \Rightarrow x^{2} - 18 x + 72 = 0 \Rightarrow x^{2} - (12 + 6) x + 72 = 0 \Rightarrow x^{2} - 12 x - 6 x + 72 = 0 \Rightarrow x (x - 12) - 6 (x - 12) = 0 \Rightarrow (x - 12) (x - 6) = 0 \Rightarrow x - 12 = 0 or x - 6 = 0 \Rightarrow x = 12 or x = 6 If x = 12, t he numbers are 12 and 6. (∵ 18 - 12 = 6) If x = 6, the numbers are 6 and12 (∵ 18 - 6 = 12) Hence, the numbers are 12 and 6.$

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Question 4:

The difference of two numbers is 5 and the difference of their reciprocals is $(\frac{1}{10})$ . Find the numbers.

Answer:

$Let the numbers be x and (x + 5) . According to the question: \frac{1}{x} - \frac{1}{x + 5} = \frac{1}{10} \Rightarrow \frac{x + 5 - x}{x (x + 5)} = \frac{1}{10} \Rightarrow \frac{5}{x^{2} + 5 x} = \frac{1}{10} \Rightarrow x^{2} + 5 x = 50 \Rightarrow x^{2} + 5 x - 50 = 0 \Rightarrow x (x + 10) - 5 (x + 10) = 0 \Rightarrow (x + 10) (x - 5) = 0 \Rightarrow x + 10 = 0 or x - 5 = 0 \Rightarrow x = - 10 or x = 5 If x = 5, the numbers are 5 and 10. (∵ 5 + 5 = 10) If x = - 10, the numbers are -10 and -5. (∵ -10 + 5 = -5) Hence, the numbers are (5, 10) and (-10, -5).$

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Question 5:

The sum of a number and its reciprocal is $3 \frac{41}{80}$ . Find the number.

Answer:

$Let the number be x . According to the question : x + \frac{1}{x} = 3 \frac{41}{80} \Rightarrow \frac{x^{2} + 1}{x} = \frac{281}{80} \Rightarrow 80 x^{2} + 80 = 281 x \Rightarrow 80 x^{2} - 281 x + 80 = 0 \Rightarrow 80 x^{2} - (256 + 25) x + 80 = 0 \Rightarrow 80 x^{2} - 256 x - 25 x + 80 = 0 \Rightarrow 16 x (5 x - 16) - 5 (5 x - 16) = 0 \Rightarrow (5 x - 16) (16 x - 5) = 0 \Rightarrow 5 x - 16 = 0 or 16 x - 5 = 0 \Rightarrow x = \frac{16}{5} or x = \frac{5}{16} Hence, the number is either \frac{16}{5} or \frac{5}{16} .$

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Question 6:

Divide 57 into two parts whose product is 782.

Answer:

$Let the two parts be x and (57 - x) . According to the question: x (57 - x) = 782 \Rightarrow 57 x - x^{2} = 782 \Rightarrow x^{2} - 57 x + 782 = 0 \Rightarrow x^{2} - (34 + 23) x + 782 = 0 \Rightarrow x^{2} - 34 x - 23 x + 782 = 0 \Rightarrow x (x - 34) - 23 (x - 34) = 0 \Rightarrow (x - 34) (x - 23) = 0 \Rightarrow x - 34 = 0 or x - 23 = 0 \Rightarrow x = 34 or x = 23 If x = 34, the t wo parts are 33 and 23. {∵ (57 - 34) = 23} If x = 23, the two parts are 24 and 34. {∵ (57 - 23) = 34} Hence, the two parts are 34 and 23.$

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Question 7:

Find two consecutive positive multiples of 3 whose product is 270.

Answer:

$Let the two consecutive multiples of 3 be 3 x and 3 (x + 1) . According to the question: 3 x × 3 (x + 1) = 270 \Rightarrow 9 x (x + 1) = 270 \Rightarrow x (x + 1) = 30 \Rightarrow x^{2} + x - 30 = 0 \Rightarrow x^{2} + (6 - 5) x - 30 = 0 \Rightarrow x^{2} + 6 x - 5 x - 30 = 0 \Rightarrow x (x + 6) - 5 (x + 6) = 0 \Rightarrow (x + 6) (x - 5) = 0 \Rightarrow x + 6 = 0 or x - 5 = 0 \Rightarrow x = - 6 or x = 5 If x = - 6, the two consecutive multiples of 3 are {3 \times (- 6) = - 18} and {3 \times (- 6) + 3 = - 15} . If x = 5, the two consecutive multiples of 3 are {3 \times 5 = 15} and {3 \times 5 + 3 = 18} Hence, the two consecutive positive multiples of 3 are (15, 18) .$

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Question 8:

Find two consecutive positive even integers, the sum of whose squares is 340.

Answer:

$Let the two consecutive even positive integers be 2 x and (2 x + 2) . According to the question: (2 x)^{2} + {(2 x + 2)}^{2} = 340 \Rightarrow 4 x^{2} + {(2 x)}^{2} + 2 \times 2 x \times 2 + {(2)}^{2} = 340 \Rightarrow 4 x^{2} + 4 x^{2} + 8 x + 4 = 340 \Rightarrow 8 x^{2} + 8 x - 336 = 0 \Rightarrow 8 (x^{2} + x - 42) = 0 \Rightarrow x^{2} + x - 42 = 0 \Rightarrow x^{2} + (7 - 6) x - 42 = 0 \Rightarrow x^{2} + 7 x - 6 x - 42 = 0 \Rightarrow x (x + 7) - 6 (x + 7) = 0 \Rightarrow (x + 7) (x - 6) = 0 \Rightarrow x + 7 = 0 or x - 6 = 0 \Rightarrow x = - 7 or x = 6 \Rightarrow x = 6 (∵ x cannot be negative) Hence, the two consecutive numbers are {(2 \times 6) = 12} and {(2 \times 6 + 2) = 14} .$

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Question 9:

The sum of a number and its squares is $\frac{63}{4}$ . Find the number.

Answer:

$Let the required number be x . According to the question: x + x^{2} = \frac{63}{4} \Rightarrow 4 x + 4 x^{2} = 63 \Rightarrow 4 x^{2} + 4 x - 63 = 0 \Rightarrow 4 x^{2} + (18 - 14) x - 63 = 0 \Rightarrow 4 x^{2} + 18 x - 14 x - 63 = 0 \Rightarrow 2 x (2 x + 9) - 7 (2 x + 9) = 0 \Rightarrow (2 x + 9) (2 x - 7) = 0 \Rightarrow 2 x + 9 = 0 or 2 x - 7 = 0 \Rightarrow x = \frac{- 9}{2} or x = \frac{7}{2} Hence, the number is either \frac{- 9}{2} or \frac{7}{2} .$

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Question 10:

The sum of a number and its positive square root is $\frac{6}{25}$ . Find the number.

Answer:

$\begin{array}{l} Let the required number be x . \\ According to the question: \end{array} x + \sqrt{x} = \frac{6}{25} \Rightarrow 25 x + 25 \sqrt{x} = 6 \Rightarrow 25 x + 25 \sqrt{x} - 6 = 0 Let \sqrt{x} be equal to y . ∴ 25 y^{2} + 25 y - 6 = 0 \Rightarrow 25 y^{2} + (30 - 5) y - 6 = 0 \Rightarrow 25 y^{2} + 30 y - 5 y - 6 = 0 \Rightarrow 5 y (5 y + 6) - 1 (5 y + 6) = 0 \Rightarrow (5 y + 6) (5 y - 1) = 0 \Rightarrow 5 y + 6 = 0 or 5 y - 1 = 0 \Rightarrow y = \frac{- 6}{5} or y = \frac{1}{5} \Rightarrow y = \frac{1}{5} (∵ y cannot be negative) \Rightarrow \sqrt{x} = \frac{1}{5} \Rightarrow x = \frac{1}{25} Hence, the required number is \frac{1}{25} .$

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Question 11:

Two natural numbers differ by 3 and their product is 504. Find the numbers.

Answer:

$Let the required numbers be x and (x + 3) . According to the question: x (x + 3) = 504 \Rightarrow x^{2} + 3 x = 504 \Rightarrow x^{2} + 3 x - 504 = 0 \Rightarrow x^{2} + (24 - 21) x - 504 = 0 \Rightarrow x^{2} + 24 x - 21 x - 504 = 0 \Rightarrow x (x + 24) - 21 (x + 24) = 0 \Rightarrow (x + 24) (x - 21) = 0 \Rightarrow x + 24 = 0 or x - 21 = 0 \Rightarrow x = - 24 or x = 21 If x = - 24, the numbers are - 24 and {(- 24 + 3) = - 21}. If x = 21, the numbers are 21 and {(21 + 3) = 24}. Hence, the numbers are (- 24, - 21) and (21, 24) .$

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Question 12:

Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

$Let the two positive integers be x and (x + 1) . According to the question: x^{2} + {(x + 1)}^{2} = 365 \Rightarrow x^{2} + x^{2} + 2 x + 1 = 365 \Rightarrow 2 x^{2} + 2 x - 364 = 0 \Rightarrow 2 (x^{2} + x - 182) = 0 \Rightarrow x^{2} + x - 182 = 0 \Rightarrow x^{2} + (14 - 13) x - 182 = 0 \Rightarrow x^{2} + 14 x - 13 x - 182 = 0 \Rightarrow x (x + 14) - 13 (x + 14) = 0 \Rightarrow (x + 14) (x - 13) = 0 \Rightarrow x + 14 = 0 or x - 13 = 0 \Rightarrow x = - 14 or x = 13 \Rightarrow x = 13 (∵ x is a positive integer) Hence, the numbers are 13 and (13 + 1 = 14).$

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Question 13:

The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.

Answer:

$Let the greater number be x and the smaller number be y . According to the question: x^{2} - y^{2} = 45 . . . (i) y^{2} = 4 x . . . (ii) From (i) and (ii), we get: x^{2} - 4 x = 45 \Rightarrow x^{2} - 4 x - 45 = 0 \Rightarrow x^{2} - (9 - 5) x - 45 = 0 \Rightarrow x^{2} - 9 x + 5 x - 45 = 0 \Rightarrow x (x - 9) + 5 (x - 9) = 0 \Rightarrow (x - 9) (x + 5) = 0 \Rightarrow x - 9 = 0 or x + 5 = 0 \Rightarrow x = 9 or x = - 5 \Rightarrow x = 9 (∵ x is a natural number) Putting the value of x in equation (ii), we get : y^{2} = 4 \times 9 \Rightarrow y^{2} = 36 \Rightarrow y = 6 Hence, the two numbers are 9 and 6 .$

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Question 14:

Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Answer:

$Let the two natural numbers be x and y . According to the question: x^{2} + y^{2} = 25 (x + y) . . . (i) x^{2} + y^{2} = 50 (x - y) . . . (ii) From (i) and (ii), we get: 25 (x + y) = 50 (x - y) \Rightarrow x + y = 2 (x - y) \Rightarrow x + y = 2 x - 2 y \Rightarrow y + 2 y = 2 x - x \Rightarrow 3 y = x . . . (iii) From (ii) and (iii), we get : {(3 y)}^{2} + y^{2} = 50 (3 y - y) \Rightarrow 9 y^{2} + y^{2} = 100 y \Rightarrow 10 y^{2} = 100 y \Rightarrow y = 10 From (iii), we have : 3 \times 10 = x \Rightarrow 30 = x Hence, the two natural numbers are 30 and 10 .$

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Question 15:

Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.

Answer:

$Let the larger and smaller parts be x and y, respectively . According to the question: x + y = 16 ... (i) 2 x^{2} = y^{2} + 164 ... (ii) From (i), we get: x = 16 - y ... (iii) From (ii) and (iii), we get: 2 {(16 - y)}^{2} = y^{2} + 164 \Rightarrow 2 (256 - 32 y + y^{2}) = y^{2} + 164 \Rightarrow 512 - 64 y + 2 y^{2} = y^{2} + 164 \Rightarrow y^{2} - 64 y + 348 = 0 \Rightarrow y^{2} - (58 + 6) y + 348 = 0 \Rightarrow y^{2} - 58 y - 6 y + 348 = 0 \Rightarrow y (y - 58) - 6 (y - 58) = 0 \Rightarrow (y - 58) (y - 6) = 0 \Rightarrow y - 58 = 0 or y - 6 = 0 \Rightarrow y = 6 (∵ y < 16) Putting the value of y in equation (iii), we get : x = 16 - 6 = 10 Hence, the two natural numbers are 6 and 10 .$

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Question 16:

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2 \frac{9}{10}$ . Find the fraction.

Answer:

$Let the numerator be x . ∴ Denominator = x + 3 ∴ Original number = \frac{x}{x + 3} According to the question: \frac{x}{x + 3} + \frac{1}{(\frac{x}{x + 3})} = 2 \frac{9}{10} \Rightarrow \frac{x}{x + 3} + \frac{x + 3}{x} = \frac{29}{10} \Rightarrow \frac{x^{2} + {(x + 3)}^{2}}{x (x + 3)} = \frac{29}{10} \Rightarrow \frac{x^{2} + x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow \frac{2 x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow 29 x^{2} + 87 x = 20 x^{2} + 60 x + 90 \Rightarrow 9 x^{2} + 27 x - 90 = 0 \Rightarrow 9 (x^{2} + 3 x - 10) = 0 \Rightarrow x^{2} + 3 x - 10 = 0 \Rightarrow x^{2} + 5 x - 2 x - 10 = 0 \Rightarrow x (x + 5) - 2 (x + 5) = 0 \Rightarrow (x - 2) (x + 5) = 0 \Rightarrow x - 2 = 0 or x + 5 = 0 \Rightarrow x = 2 or x = - 5 (rejected) So, numerator = x = 2 denominator = x + 3 = 2 + 3 = 5 So, required fraction = \frac{2}{5}$

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Question 17:

A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ O riginal number = 10 y + x According to the question: 10 y + x = 4 (x + y) \Rightarrow 10 y + x = 4 x + 4 y \Rightarrow 3 x - 6 y = 0 \Rightarrow 3 x = 6 y \Rightarrow x = 2 y .... (i) Also, 10 y + x = 2 x y \Rightarrow 10 y + 2 y = 2 .2 y . y [From (i)] \Rightarrow 12 y = 4 y^{2} \Rightarrow y = 3 From (i), we get: x = 2 \times 3 = 6 ∴ Original number = 10 \times 3 + 6 = 36$

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Question 18:

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ x y = 14 ⇒ y = \frac{14}{x} ... (i) According to the question: (10 y + x) + 45 = 10 x + y \Rightarrow 9 y - 9 x = - 45 \Rightarrow y - x = - 5 ... (ii) From (i) and (ii), we get: \frac{14}{x} - x = - 5 \Rightarrow \frac{14 - x^{2}}{x} = - 5 \Rightarrow 14 - x^{2} = - 5 x \Rightarrow x^{2} - 5 x - 14 = 0 \Rightarrow x^{2} - (7 - 2) x - 14 = 0 \Rightarrow x^{2} - 7 x + 2 x - 14 = 0 \Rightarrow x (x - 7) + 2 (x - 7) = 0 \Rightarrow (x - 7) (x + 2) = 0 \Rightarrow x - 7 = 0 or x + 2 = 0 \Rightarrow x = 7 or x = - 2 \Rightarrow x = 7 (∵ the digit cannot be negative) Putting x = 7 in equation (i), we get: y = 2 ∴ Required number = 10 \times 2 + 7 = 27$

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Question 19:

Out of a number of saras birds, one-fourth of the number are moving about in lots, $\frac{1}{9} th$ coupled with $\frac{1}{4} th$ as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula trees. What is the total number of birds?

Answer:

$Let the total number of birds be x^{2} . According to the question: \frac{x^{2}}{4} + (\frac{x^{2}}{9} + \frac{x^{2}}{4}) + 7 x + 56 = x^{2} \Rightarrow \frac{9 x^{2} + 4 x^{2} + 9 x^{2} + 252 x + 2016}{36} = x^{2} \Rightarrow 22 x^{2} + 252 x + 2016 = 36 x^{2} \Rightarrow 14 x^{2} - 252 x - 2016 = 0 \Rightarrow x^{2} - 18 x - 144 = 0 \Rightarrow x^{2} - 24 x + 6 x - 144 = 0 \Rightarrow x (x - 24) + 6 (x - 24) = 0 \Rightarrow (x - 24) (x + 6) = 0 \Rightarrow x = 24 or x = - 6 ∴ x = 24 (∵ N umber of birds cannot be negative) ∴ N umber of birds = x^{2} = 576$

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Question 20:

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.

Answer:

$Let there be x rows. Then, the number of students in each row will also be x . ∴ T otal number of students = (x^{2} + 24) According to the question: {(x + 1)}^{2} - 25 = x^{2} + 24 \Rightarrow x^{2} + 2 x + 1 - 25 - x^{2} - 24 = 0 \Rightarrow 2 x - 48 = 0 \Rightarrow 2 x = 48 \Rightarrow x = 24 ∴ {Total number of students = 24}^{2} + 24 = 576 + 24 = 600$

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Question 21:

300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of student.

Answer:

$\begin{array}{l} Let the total number of students be x . \end{array} According to the question : \frac{300}{x} - \frac{300}{x + 10} = 1 \Rightarrow \frac{300 (x + 10) - 300 x}{x (x + 10)} = 1 \Rightarrow \frac{300 x + 3000 - 300 x}{x^{2} + 10 x} = 1 \Rightarrow 3000 = x^{2} + 10 x \Rightarrow x^{2} + 10 x - 3000 = 0 \Rightarrow x^{2} + (60 - 50) x - 3000 = 0 \Rightarrow x^{2} + 60 x - 50 x - 3000 = 0 \Rightarrow x (x + 60) - 50 (x + 60) = 0 \Rightarrow (x + 60) (x - 50) = 0 \Rightarrow x = 50 or x = - 60 x cannot be negative; therefore, the total number of students is 50.$

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Question 22:

A man busy a number of pens for Rs 80. If he had bought 4 more pens for the same amount, each pen would have cost him Rs 1 less. How many pens did he buy?

Answer:

$\begin{array}{l} Let the total number of pens be x . \end{array} According to the question : \frac{80}{x} - \frac{80}{x + 4} = 1 \Rightarrow \frac{80 (x + 4) - 80 x}{x (x + 4)} = 1 \Rightarrow \frac{80 x + 320 - 80 x}{x^{2} + 4 x} = 1 \Rightarrow 320 = x^{2} + 4 x \Rightarrow x^{2} + 4 x - 320 = 0 \Rightarrow x^{2} + (20 - 16) x - 320 = 0 \Rightarrow x^{2} + 20 x - 16 x - 320 = 0 \Rightarrow x (x + 20) - 16 (x + 20) = 0 \Rightarrow (x + 20) (x - 16) = 0 \Rightarrow x = - 20 or x = 16 T he total number of pens cannot be negative; therefore, the total number of pens is 16.$

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Question 23:

In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.

Answer:

$\begin{array}{l} Let the marks of Kamal in mathematics and english be x and y, respectively. \end{array} According to the question : x + y = 40 . . . (i) Also, (x + 3) (y - 4) = 360 \Rightarrow (x + 3) (40 - x - 4) = 360 [F rom (i)] \Rightarrow (x + 3) (36 - x) = 360 \Rightarrow 36 x - x^{2} + 108 - 3 x = 360 \Rightarrow 33 x - x^{2} - 252 = 0 \Rightarrow - x^{2} + 33 x - 252 = 0 \Rightarrow x^{2} - 33 x + 252 = 0 \Rightarrow x^{2} - (21 + 12) x + 252 = 0 \Rightarrow x^{2} - 21 x - 12 x + 252 = 0 \Rightarrow x (x - 21) - 12 (x - 21) = 0 \Rightarrow (x - 21) (x - 12) = 0 \Rightarrow x = 21 or x = 12 If x = 21, y = 40 - 21 = 19 Thus, Kamal scored 21 and 19 m arks in mathematics and english, respectively. If x = 12, y = 40 - 12 = 28 Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.$

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Question 24:

A takes 10 days than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

Answer:

$\begin{array}{l} Let B takes x days to complete the work. \end{array} Therefore, A will take (x - 10) days. ∴ \frac{1}{x} + \frac{1}{(x - 10)} = \frac{1}{12} \Rightarrow \frac{(x - 10) + x}{x (x - 10)} = \frac{1}{12} \Rightarrow \frac{2 x - 10}{x^{2} - 10 x} = \frac{1}{12} \Rightarrow x^{2} - 10 x = 12 (2 x - 10) \Rightarrow x^{2} - 10 x = 24 x - 120 \Rightarrow x^{2} - 34 x + 120 = 0 \Rightarrow x^{2} - (30 + 4) x + 120 = 0 \Rightarrow x^{2} - 30 x - 4 x + 120 = 0 \Rightarrow x (x - 30) - 4 (x - 30) = 0 \Rightarrow (x - 30) (x - 4) = 0 \Rightarrow x = 30 or x = 4 Number of days to complete the work by B cannot be less than that by A; therefore, we get: x = 30 Thus, B completes the work in 30 days .$

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Question 25:

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

Answer:

$\begin{array}{l} Let the usual speed be x km/hr . \end{array} According to the question : \frac{300}{x} - \frac{300}{(x + 5)} = 2 \Rightarrow \frac{300 (x + 5) - 300 x}{x (x + 5)} = 2 \Rightarrow \frac{300 x + 1500 - 300 x}{x^{2} + 5 x} = 2 \Rightarrow 1500 = 2 (x^{2} + 5 x) \Rightarrow 1500 = 2 x^{2} + 10 x \Rightarrow x^{2} + 5 x - 750 = 0 \Rightarrow x^{2} + (30 - 25) x - 750 = 0 \Rightarrow x^{2} + 30 x - 25 x - 750 = 0 \Rightarrow x (x + 30) - 25 (x + 30) = 0 \Rightarrow (x + 30) (x - 25) = 0 \Rightarrow x = - 30 or x = 25 T he usual speed cannot be negative; therefore, the speed is 25 km/hr.$

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Question 26:

A train travels a distance of 360 km at a uniform speed. If the speed of he train is increased by 5 km an hour, the journey would have taken 1 hour less. Find the original speed of the train.

Answer:

$\begin{array}{l} Let the original speed of the train be x km/hr . \end{array} According to the question : \frac{360}{x} - \frac{360}{(x + 5)} = 1 \Rightarrow \frac{360 (x + 5) - 360 x}{x (x + 5)} = 1 \Rightarrow \frac{360 x + 1800 - 360 x}{x^{2} + 5 x} = 1 \Rightarrow 1800 = x^{2} + 5 x \Rightarrow x^{2} + 5 x - 1800 = 0 \Rightarrow x^{2} + (45 - 40) x - 1800 = 0 \Rightarrow x^{2} + 45 x - 40 x - 1800 = 0 \Rightarrow x (x + 45) - 40 (x + 45) = 0 \Rightarrow (x + 45) (x - 40) = 0 \Rightarrow x = - 45 or x = 40 Th e speed of train cannot be negative; therefore, the speed is 40 km/hr.$

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Question 27:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Answer:

$\begin{array}{l} Let the original speed of the train be x km/hr . \end{array} According to the question : \frac{90}{x} - \frac{90}{(x + 15)} = \frac{1}{2} \Rightarrow \frac{90 (x + 15) - 90 x}{x (x + 15)} = \frac{1}{2} \Rightarrow \frac{90 x + 1350 - 90 x}{x^{2} + 15 x} = \frac{1}{2} \Rightarrow \frac{1350}{x^{2} + 15 x} = \frac{1}{2} \Rightarrow 2700 = x^{2} + 15 x \Rightarrow x^{2} + (60 - 45) x - 2700 = 0 \Rightarrow x^{2} + 60 x - 45 x - 2700 = 0 \Rightarrow x (x + 60) - 45 (x + 60) = 0 \Rightarrow (x + 60) (x - 45) = 0 \Rightarrow x = - 60 or x = 45 x cannot be negative; therefore, the original speed of train is 45 km/hr.$

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Question 28:

The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by 20 km/hr.

Answer:

$\begin{array}{l} Let the speed of the Deccan Queen be x km/hr . \end{array} According to the question : Speed of another train = (x - 20) km / h ∴ \frac{192}{x - 20} - \frac{192}{x} = \frac{48}{60} \Rightarrow \frac{4}{x - 20} - \frac{4}{x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 (x - 20)}{(x - 20) x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 x + 80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow \frac{80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow x^{2} - 20 x = 4800 \Rightarrow x^{2} - 20 x - 4800 = 0 \Rightarrow x^{2} - (80 - 60) x - 4800 = 0 \Rightarrow x^{2} - 80 x + 60 x - 4800 = 0 \Rightarrow x (x - 80) + 60 (x - 80) = 0 \Rightarrow (x - 80) (x + 60) = 0 \Rightarrow x = 80 or x = - 60 The value of x cannot be negative; therefore, the original speed of Deccan Queen is 80 km/hr.$

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Question 29:

A motorboat whose speed is 9 km/hr in still water, goes 15 km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Answer:

$\begin{array}{l} Let the speed of the stream be x km / hr . \end{array} ∴ D ownstream speed= (9 + x) km/hr Upstream speed= (9 - x) km/hr Distance covered downstream = Distance covered upstream = 15 km Total time taken = 3 hours 45 minutes = (3 + \frac{45}{60}) minutes = \frac{225}{60} minutes = \frac{15}{4} minutes ∴ \frac{15}{(9 + x)} + \frac{15}{(9 - x)} = \frac{15}{4} \Rightarrow \frac{1}{(9 + x)} + \frac{1}{(9 - x)} = \frac{1}{4} \Rightarrow \frac{9 - x + 9 + x}{(9 + x) (9 - x)} = \frac{1}{4} \Rightarrow \frac{18}{9^{2} - x^{2}} = \frac{1}{4} \Rightarrow \frac{18}{81 - x^{2}} = \frac{1}{4} \Rightarrow 81 - x^{2} = 72 \Rightarrow 81 - x^{2} - 72 = 0 \Rightarrow - x^{2} + 9 = 0 \Rightarrow x^{2} = 9 \Rightarrow x = 3 or x = - 3 The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.$

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Question 30:

A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than o return to the same spot. Find the speed of the stream.

Answer:

$\begin{array}{l} Let the speed of the stream be x km/hr. \end{array} Given : Speed of the boat = 18 km / hr ∴ Speed downstream = (18 + x) km/hr S peed upstream = (18 - x) km/hr ∴ \frac{24}{(18 - x)} - \frac{24}{(18 + x)} = 1 \Rightarrow \frac{1}{(18 - x)} - \frac{1}{(18 + x)} = \frac{1}{24} \Rightarrow \frac{18 + x - 18 + x}{(18 - x) (18 + x)} = \frac{1}{24} \Rightarrow \frac{2 x}{18^{2} - x^{2}} = \frac{1}{24} \Rightarrow 324 - x^{2} = 48 x \Rightarrow 324 - x^{2} - 48 x = 0 \Rightarrow x^{2} + 48 x - 324 = 0 \Rightarrow x^{2} + (54 - 6) x - 324 = 0 \Rightarrow x^{2} + 54 x - 6 x - 324 = 0 \Rightarrow x (x + 54) - 6 (x + 54) = 0 \Rightarrow (x + 54) (x - 6) = 0 \Rightarrow x = - 54 or x = 6 The value of x cannot be negative; thereore, the speed of the stream is 6 km/hr.$

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Question 31:

The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Answer:

$Speed of the boat in still water = 8 km/hr Let the speed of the stream be x km/hr. ∴ Speed upstream = (8 - x) km/hr Speed downstream = (8 + x) km/hr Time taken to go 22 km downstream = \frac{22}{(8 + x)} hr Time taken to go 15 km upstream = \frac{15}{(8 - x)} hr According to the question : \Rightarrow \frac{22}{(8 + x)} + \frac{15}{(8 - x)} = 5 \Rightarrow \frac{22}{(8 + x)} + \frac{15}{(8 - x)} - 5 = 0 \Rightarrow \frac{22 (8 - x) + 15 (8 + x) - 5 (8 - x) (8 + x)}{(8 - x) (8 + x)} = 0 \Rightarrow 176 - 22 x + 120 + 15 x - 320 + 5 x^{2} = 0 \Rightarrow 5 x^{2} - 7 x - 24 = 0 \Rightarrow 5 x^{2} - (15 - 8) x - 24 = 0 \Rightarrow 5 x^{2} - 15 x + 8 x - 24 = 0 \Rightarrow 5 x (x - 3) - 8 (x - 3) = 0 \Rightarrow (x - 3) (5 x - 8) = 0 \Rightarrow x - 3 = 0 or 5 x - 8 = 0 \Rightarrow x = 3 or x = \frac{8}{5} \Rightarrow x = 3 (∵ Speed cannot be a fraction) ∴ Speed of the stream = 3 km/hr$

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Question 32:

A sailor can row a boat 8 km downstream and return back to the starting point in 1 hour 40 minutes. If the speed of the stream is 2 kmph, find the speed of the boat in still water.

Answer:

$Let the speed of the boat in still water be x km/hr. It is given that the speed of the stream is 2 km/hr. ∴ S peed upstream = (x - 2) km/hr Speed downstream = (x + 2) km/hr Time taken to go 8 km downstream = \frac{8}{(x + 2)} hr Time taken to go 8 km upstream = \frac{8}{(x - 2)} hr ∴ \frac{8}{(x + 2)} + \frac{8}{(x - 2)} = 1 \frac{40}{60} \Rightarrow \frac{8}{(x + 2)} + \frac{8}{(x - 2)} = \frac{100}{60} \Rightarrow \frac{8 (x - 2) + 8 (x + 2)}{(x - 2) (x + 2)} = \frac{5}{3} \Rightarrow \frac{16 x}{x^{2} - 4} = \frac{5}{3} \Rightarrow 3 \times 16 x = 5 (x^{2} - 4) \Rightarrow 48 x = 5 x^{2} - 20 \Rightarrow 5 x^{2} - 48 x - 20 = 0 \Rightarrow 5 x^{2} - (50 - 2) x - 20 = 0 \Rightarrow 5 x^{2} - 50 x + 2 x - 20 = 0 \Rightarrow 5 x (x - 10) + 2 (x - 10) = 0 \Rightarrow (x - 10) (5 x + 2) = 0 \Rightarrow x - 10 = 0 or 5 x + 2 = 0 \Rightarrow x = 10 or x = \frac{- 2}{5} \Rightarrow x = 10 (∵ Speed cannot be a negative fraction) ∴ Speed of the boat in still water = 10 km/hr$

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Question 33:

Two pipes running together can fill a cistern in $3 \frac{1}{13}$ minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Answer:

$Let one pipe fills the cistern in x mins. Therefore, the o ther pipe will fill the cistern in (x + 3) mins. Time taken by both, running together, to fill the cistern = 3 \frac{1}{13} mins = \frac{40}{13} mins Part filled by one pipe in 1 min = \frac{1}{x} Part filled by the other pipe in 1 min = \frac{1}{x + 3} Part filled by both pipes, running together, in 1 min = \frac{1}{x} + \frac{1}{x + 3} ∴ \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{\frac{40}{13}} \Rightarrow \frac{(x + 3) + x}{x (x + 3)} = \frac{13}{40} \Rightarrow \frac{2 x + 3}{x^{2} + 3 x} = \frac{13}{40} \Rightarrow 13 x^{2} + 39 x = 80 x + 120 \Rightarrow 13 x^{2} - 41 x - 120 = 0 \Rightarrow 13 x^{2} - (65 - 24) x - 120 = 0 \Rightarrow 13 x^{2} - 65 x + 24 x - 120 = 0 \Rightarrow 13 x (x - 5) + 24 (x - 5) = 0 \Rightarrow (x - 5) (13 x + 24) = 0 \Rightarrow x - 5 = 0 or 13 x + 24 = 0 ⇒ x = 5 or x = \frac{- 24}{13} \Rightarrow x = 5 (∵ Speed cannot be a negative fraction) Thus, o ne pipe will take 5 mins and the other will take {(5 + 3) = 8} mins to fill the cistern .$

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Question 34:

One year ago, a man was 8 times old as his son. Now his age is equal to the square of his son's age. Find their present ages.

Answer:

$Let the present ages of the father and his son be x years and y years, respectively . According to the question : x - 1 = 8 (y - 1) \Rightarrow x - 1 = 8 y - 8 \Rightarrow x = 8 y - 7 . . . (i) Also, x = y^{2} . . . (ii) From (i) and (ii), we get: 8 y - 7 = y^{2} \Rightarrow y^{2} - 8 y + 7 = 0 ⇒ y^{2} - (7 + 1) y + 7 = 0 ⇒ y^{2} - 7 y - y + 7 = 0 ⇒ y (y - 7) - 1 (y - 7) = 0 ⇒ (y - 7) (y - 1) = 0 ⇒ y - 7 = 0 or y - 1 = 0 ⇒ y = 7 or y = 1 ⇒ y = 7 (∵ y > 1) Putting the value of y in equation (i), we get : x = 49 Thus, the present ages of the father and his son are 49 years and 7 years, respectively.$

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Question 35:

The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at that time. Find their present ages.

Answer:

$Let the present ages of the man and his son be x years and y years, respectively . According to the question : x + y = 45 \Rightarrow x = 45 - y . . . (i) Also, (x - 5) (y - 5) = 4 (x - 5) . . . (ii) From (i) and (ii), we get: (45 - y - 5) (y - 5) = 4 (45 - y - 5) \Rightarrow (40 - y) (y - 5) = 4 (40 - y) \Rightarrow 40 y - 200 - y^{2} + 5 y = 160 - 4 y \Rightarrow y^{2} - 49 y + 360 = 0 \Rightarrow y^{2} - (40 + 9) y + 360 = 0 \Rightarrow y^{2} - 40 y - 9 y + 360 = 0 \Rightarrow y (y - 40) - 9 (y - 40) = 0 \Rightarrow (y - 9) (y - 40) = 0 \Rightarrow y - 9 = 0 [∵ x + y = 45, x > y] \Rightarrow y = 9 Putting the value of y in equation (i), we get : x = 36 Hence, the present ages of the man and his son are 36 years and 9 years, respectively.$

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Question 36:

The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age.

Answer:

$Let the present age of Meena be x years . According to the question : (x - 5) (x + 8) = 30 \Rightarrow x^{2} + 3 x - 40 = 30 \Rightarrow x^{2} + 3 x - 70 = 0 \Rightarrow x^{2} + (10 - 7) x - 70 = 0 \Rightarrow x^{2} + 10 x - 7 x - 70 = 0 \Rightarrow x (x + 10) - 7 (x + 10) = 0 \Rightarrow (x + 10) (x - 7) = 0 \Rightarrow x + 10 = 0 or x - 7 = 0 \Rightarrow x = - 10 or x = 7 \Rightarrow x = 7 (∵ Age cannot be negative) Thus, the present age of Meena is 7 years .$

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Question 37:

The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.

Answer:

$Let the present ages of the boy and his brother be x years and (25 - x) years . According to the question : x (25 - x) = 126 \Rightarrow 25 x - x^{2} = 126 \Rightarrow x^{2} - (18 + 7) x + 126 = 0 \Rightarrow x^{2} - 18 x - 7 x + 126 = 0 \Rightarrow x (x - 18) - 7 (x - 18) = 0 \Rightarrow (x - 18) (x - 7) = 0 \Rightarrow x - 18 = 0 or x - 7 = 0 \Rightarrow x = 18 or x = 7 \Rightarrow x = 18 (∵ P resent age of the boy cannot be less than his brother) If x = 18, we have : Present ages of the boy = 18 years Present age of his brother = (25 - 18) years = 7 years Thus, the present ages of the boy and his brother are 18 years and 7 years, respectively.$

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Question 38:

A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m². Find the width of the path.

Answer:

$Let the width of the path be x m. ∴ Length of the field including the path = 16 + x + x = 16 + 2 x Breadth of the field including the path = 10 + x + x = 10 + 2 x Now, (Area of the field including path) - (A rea of the field excluding path) = Area of the path \Rightarrow (16 + 2 x) (10 + 2 x) - (16 \times 10) = 120 \Rightarrow 160 + 32 x + 20 x + 4 x^{2} - 160 = 120 \Rightarrow 4 x^{2} + 52 x - 120 = 0 \Rightarrow x^{2} + 13 x - 30 = 0 \Rightarrow x^{2} + (15 - 2) x + 30 = 0 \Rightarrow x^{2} + 15 x - 2 x + 30 = 0 \Rightarrow x (x + 15) - 2 (x + 15) = 0 \Rightarrow (x - 2) (x + 15) = 0 \Rightarrow x - 2 = 0 or x + 15 = 0 \Rightarrow x = 2 or x = - 15 \Rightarrow x = 2 (∵ Width cannot be negative) Thus, the width of the path is 2 m.$

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Question 39:

The length of a rectangle is twice its breadth and its area is 288 cm². Find the dimensions of the rectangle.

Answer:

$Let the length and breadth of the rectangle be 2 x m and x m, respectively. According to the question : 2 x \times x = 288 \Rightarrow 2 x^{2} = 288 \Rightarrow x^{2} = 144 \Rightarrow x = 12 or x = - 12 \Rightarrow x = 12 (∵ x cannot be negative) ∴ L ength = 2 \times 12 = 24 m Breadth = 12 m$

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Question 40:

The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.

Answer:

$Let the length and breadth of the rectangle be 3 x m and x m, respectively. According to the question : 3 x \times x = 147 \Rightarrow 3 x^{2} = 147 \Rightarrow x^{2} = 49 \Rightarrow x = 7 or x = - 7 \Rightarrow x = 7 (∵ x cannot be negative) ∴ Length = 3 \times 7 = 21m Breadth = 7 m$

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Question 41:

The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.

Answer:

$\begin{array}{l} Let the breadth of the rectangular hall be x metre. \end{array} Therefore, the length of the rectangular hall will be (x + 3) metre. According to the question : x (x + 3) = 238 \Rightarrow x^{2} + 3 x = 238 \Rightarrow x^{2} + 3 x - 238 = 0 \Rightarrow x^{2} + (17 - 14) x - 238 = 0 \Rightarrow x^{2} + 17 x - 14 x - 238 = 0 \Rightarrow x (x + 17) - 14 (x + 17) = 0 \Rightarrow (x + 17) (x - 14) = 0 \Rightarrow x = - 17 or x = 14 But the value x cannot be negative. Therefore, the breadth of the hall is 14 metre and the length is 17 metre.$

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Question 42:

The perimeter of a rectangular plot is 62 m and its area is 228 sq metres. Find the dimensions of the plot.

Answer:

$Let the length and breadth of the rectangular plot be x and y meter, respectively . Therefore, we have : Perimeter = 2 (x + y) = 62 . . . (i) and A rea = x y = 228 \Rightarrow y = \frac{228}{x} Putting the value of y in (i), we get ⇒ 2 (x + \frac{228}{x}) = 62 \Rightarrow x + \frac{228}{x} = 31 \Rightarrow \frac{x^{2} + 228}{x} = 31 \Rightarrow x^{2} + 228 = 31 x \Rightarrow x^{2} - 31 x + 228 = 0 \Rightarrow x^{2} - (19 + 12) x + 228 = 0 \Rightarrow x^{2} - 19 x - 12 x + 228 = 0 \Rightarrow x (x - 19) - 12 (x - 19) = 0 \Rightarrow (x - 19) (x - 12) = 0 \Rightarrow x = 19 or x = 12 If x = 19 m, y = \frac{228}{19} = 12 m Therefore, the length and breadth of the plot are 19 m and 12 m, respectively .$

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Question 43:

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than width of a the rectangle. Their areas being equal, find their dimensions.

Answer:

$Let the breadth of rectangle be x cm . According to the question : Side of the square = (x + 4) cm Length of the rectangle = {3 (x + 4)} cm It is given that the areas of the rectangle and square are same. ∴ 3 (x + 4) \times x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = x^{2} + 8 x + 16 \Rightarrow 2 x^{2} + 4 x - 16 = 0 \Rightarrow x^{2} + 2 x - 8 = 0 \Rightarrow x^{2} + (4 - 2) x - 8 = 0 \Rightarrow x^{2} + 4 x - 2 x - 8 = 0 \Rightarrow x (x + 4) - 2 (x + 4) = 0 \Rightarrow (x + 4) (x - 2) = 0 \Rightarrow x = - 4 or x = 2 ∴ x = 2 (∵ The value of x cannot be negative) Thus, t he breadth of the rectangle is 2 cm and length is {3 (2 + 4) =18} cm. Also, the side of the square is 6 cm.$

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Question 44:

A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

Answer:

$Let the length and breadth of the rectangular garden be x and y metre, respectively. Given: x y = 180 sq m ... (i) and 2 y + x = 39 \Rightarrow x = 39 - 2 y Putting the value of x in (i), we get: (39 - 2 y) y = 180 \Rightarrow 39 y - 2 y^{2} = 180 \Rightarrow 39 y - 2 y^{2} - 180 = 0 \Rightarrow 2 y^{2} - 39 y + 180 = 0 \Rightarrow 2 y^{2} - (24 + 15) y + 180 = 0 \Rightarrow 2 y^{2} - 24 y - 15 y + 180 = 0 \Rightarrow 2 y (y - 12) - 15 (y - 12) = 0 \Rightarrow (y - 12) (2 y - 15) = 0 \Rightarrow y = 12 or y = \frac{15}{2} = 7.5 If y = 12, x = 39 - 24 = 15 If y = 7.5, x = 39 - 15 = 24 Thus, the l ength and breadth of the garden are (15 m and 12 m) or (24 m and 7 .5 m), respectively.$

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Question 45:

The area of a right-triangle is 600 cm². If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.
$Area of triangle = \frac{1}{2} x (x + 10) = 600 \Rightarrow x (x + 10) = 1200 \Rightarrow x^{2} + 10 x - 1200 = 0 \Rightarrow x^{2} + (40 - 30) x - 1200 = 0 \Rightarrow x^{2} + 40 x - 30 x - 1200 = 0 \Rightarrow x (x + 40) - 30 (x + 40) = 0 \Rightarrow (x + 40) (x - 30) = 0 \Rightarrow x = - 40 or x = 30 \Rightarrow x = 30 [∵ Altitude cannot be negative] Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively. ∵ {(Hypotenuse)}^{2} = {(Altitude)}^{2} + {(Base)}^{2} \Rightarrow {(Hypotenuse)}^{2} = {(30)}^{2} + {(40)}^{2} \Rightarrow {(Hypotenuse)}^{2} = 900 + 1600 = 2500 \Rightarrow {(Hypotenuse)}^{2} = {(50)}^{2} \Rightarrow (Hypotenuse) = 50 Thus, the dimensions of the triangle are :, Hypotenuse = 50 cm Altitude = 30 cm Base = 40 cm$

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Question 46:

The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.

Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times 3 x \times x = 96 (∵ Area = 96 sq m) \Rightarrow \frac{x^{2}}{2} = 32 \Rightarrow x^{2} = 64 \Rightarrow x = \pm 8$

The value of x cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 $\times$ 8 = 24 m), respectively.

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Question 47:

The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.

Answer:

Let the base be $x$ m.
Therefore, the altitude will be $(x + 7)$ m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times x \times (x + 7) = 165 \Rightarrow x^{2} + 7 x = 330 \Rightarrow x^{2} + 7 x - 330 = 0 \Rightarrow x^{2} + (22 - 15) x - 330 = 0 \Rightarrow x^{2} + 22 x - 15 x - 330 = 0 \Rightarrow x (x + 22) - 15 (x + 22) = 0 \Rightarrow (x + 22) (x - 15) = 0 \Rightarrow x = - 22 or x = 15$

The value of $x$ cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

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Question 48:

The hypotenuse of a right-angled triangle is 20 metres. If the difference between the length of the other sides be 4 metres, find the other sides.

Answer:

Let one side of the right-angled triangle be $x$ m and the other side be $(x + 4)$ m.
On applying Pythagoras theorem, we have:

$20^{2} = {(x + 4)}^{2} + x^{2} \Rightarrow 400 = x^{2} + 8 x + 16 + x^{2} \Rightarrow 2 x^{2} + 8 x - 384 = 0 \Rightarrow x^{2} + 4 x - 192 = 0 \Rightarrow x^{2} + (16 - 12) x - 192 = 0 \Rightarrow x^{2} + 16 x - 12 x - 192 = 0 \Rightarrow x (x + 16) - 12 (x + 16) = 0 \Rightarrow (x + 16) (x - 12) = 0 \Rightarrow x = - 16 or x = 12$

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

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Question 49:

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

Answer:

Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively.
Therefore, the hypotenuse will be $(x + 2)$ cm.
$∴ {(x + 2)}^{2} = y^{2} + x^{2} . . . (i) Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. ∴ h = (2 y + 1) \Rightarrow x + 2 = 2 y + 1 \Rightarrow x = 2 y - 1 P utting the value of x in (i), we get : {(2 y - 1 + 2)}^{2} = y^{2} + {(2 y - 1)}^{2} \Rightarrow {(2 y + 1)}^{2} = y^{2} + 4 y^{2} - 4 y + 1 \Rightarrow 4 y^{2} + 4 y + 1 = 5 y^{2} - 4 y + 1 \Rightarrow - y^{2} + 8 y = 0 \Rightarrow y^{2} - 8 y = 0 \Rightarrow y (y - 8) = 0 \Rightarrow y = 8 cm ∴ x = 16 - 1 = 15 cm ∴ h = 16 + 1 = 17 cm$

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

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Question 50:

The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle.

Answer:

Let the shortest side be $x$ m.
Therefore, according to the question:
Hypotenuse = $(2 x - 1)$ m
Third side = $(x + 1)$ m
On applying Pythagoras theorem, we get:

${(2 x - 1)}^{2} = {(x + 1)}^{2} + x^{2} \Rightarrow 4 x^{2} - 4 x + 1 = x^{2} + 2 x + 1 + x^{2} \Rightarrow 2 x^{2} - 6 x = 0 \Rightarrow 2 x (x - 3) = 0 \Rightarrow x = 0 or x = 3$

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = $(2 \times 3 - 1) =$ 5 m
Third side = (3 + 1) = 4 m

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Question 51:

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Answer:

Let the length of the side of the first and the second square be $x$ and y, respectively.

$According to the question : x^{2} + y^{2} = 640 . . . (i) Also, 4 x - 4 y = 64 \Rightarrow x - y = 16 \Rightarrow x = 16 + y$

Putting the value of $x$ in (i), we get:

$x^{2} + y^{2} = 640 \Rightarrow {(16 + y)}^{2} + y^{2} = 640 \Rightarrow 256 + 32 y + y^{2} + y^{2} = 640 \Rightarrow 2 y^{2} + 32 y - 384 = 0 \Rightarrow y^{2} + 16 y - 192 = 0 \Rightarrow y^{2} + (24 - 8) y - 192 = 0 \Rightarrow y^{2} + 24 y - 8 y - 192 = 0 \Rightarrow y (y + 24) - 8 (y + 24) = 0 \Rightarrow (y + 24) (y - 8) = 0 \Rightarrow y = - 24 or y = 8 ∴ y = 8 (∵ S ide cannot be negative) ∴ x = 16 + y = 16 + 8 = 24 m Thus, the sides of the squares are 8 m and 24 m.$

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Question 52:

The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.

Answer:

$Let the smaller number be x . ∴ L arger number = (2 x - 5) Also, {(2 x - 5)}^{2} - x^{2} = 88 \Rightarrow 4 x^{2} - 20 x + 25 - x^{2} = 88 \Rightarrow 3 x^{2} - 20 x - 63 = 0 \Rightarrow 3 x^{2} - (27 - 7) x - 63 = 0 \Rightarrow 3 x^{2} - 27 x + 7 x - 63 = 0 \Rightarrow 3 x (x - 9) + 7 (x - 9) = 0 \Rightarrow (x - 9) (3 x + 7) = 0 \Rightarrow x = 9 or x = \frac{7}{3} The number cannot be a fraction; therefore, we have : Smaller number = 9 Larger number = (2 × 9 - 5) = 13$

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Question 53:

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46, find the integers.

Answer:

$Let the required integers be x, (x + 1) and (x + 2) . According to the question : x^{2} + (x + 1) (x + 2) = 46 \Rightarrow x^{2} + x^{2} + 2 x + x + 2 = 46 \Rightarrow 2 x^{2} + 3 x - 44 = 0 \Rightarrow 2 x^{2} + (11 - 8) x - 44 = 0 \Rightarrow 2 x^{2} + 11 x - 8 x - 44 = 0 \Rightarrow x (2 x + 11) - 4 (2 x + 11) = 0 \Rightarrow (2 x + 11) (x - 4) = 0 \Rightarrow x = - \frac{11}{2} or x = 4 The value of x cannot be negative; therefore, we have: x = 4 Thus, the consecutive integers are 4, 5 and 6.$

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Question 1:

Which of the following is a quadratic equation?
(a) $x^{2} - 3 \sqrt{x} + 2 = 0$
(b) $x + \frac{1}{x} = x^{2}$
(c) $x^{2} + \frac{1}{x^{2}} = 5$
(d) $2 x^{2} - 5 x = (x - 1)^{2}$

Answer:

(d) $2 x^{2} - 5 x = (x - 1)^{2}$

$A quadratic equation is the equation with degree 2 . ∵ 2 x^{2} - 5 x = {(x - 1)}^{2} \Rightarrow 2 x^{2} - 5 x = x^{2} - 2 x + 1 \Rightarrow 2 x^{2} - 5 x - x^{2} + 2 x - 1 = 0 \Rightarrow x^{2} - 3 x - 1 = 0, which is a quadratic equation$

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Question 2:

Which of the following is a quadratic equation?
(a) $(x^{2} + 1) = (2 - x)^{2} + 3$
(b) $x^{3} - x^{2} = (x - 1)^{3}$
(c) $2 x^{2} + 3 = (5 + x) (2 x - 3)$
(d) None of these

Answer:

(b) $x^{3} - x^{2} = (x - 1)^{3}$

$∵ x^{3} - x^{2} = {(x - 1)}^{3} ⇒ x^{3} - x^{2} = x^{3} - 3 x^{2} + 3 x - 1 ⇒ 2 x^{2} - 3 x + 1 = 0, which is a quadratic equation$

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Question 3:

Which of the following is not a quadratic equation?
(a) $3 x - x^{2} = x^{2} + 5$
(b) $(x + 2)^{2} = 2 (x^{2} - 5)$
(c) $(\sqrt{2} x + 3)^{2} = 2 x^{2} + 6$
(d) $(x - 1)^{2} = 3 x^{2} + x - 2$

Answer:

(c) $(\sqrt{2} x + 3)^{2} = 2 x^{2} + 6$

$∵ {(\sqrt{2} x + 3)}^{2} = 2 x^{2} + 6 ⇒ 2 x^{2} + 9 + 6 \sqrt{2} x = 2 x^{2} + 6 ⇒ 6 \sqrt{2} x + 3 = 0, which is not a quadratic equation$

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Question 4:

If x = 3 is a solution of the equation $3 x^{2} + (k - 1) x + 9 = 0,$ then k = ?
(a) 11
(b) −11
(c) 13
(d) −13

Answer:

(b) −11
$It is given that x = 3 is a solution of 3 x^{2} + (k - 1) x + 9 = 0; therefore, we have : 3 {(3)}^{2} + (k - 1) \times 3 + 9 = 0 \Rightarrow 27 + 3 (k - 1) + 9 = 0 \Rightarrow 3 (k - 1) = - 36 \Rightarrow (k - 1) = - 12 \Rightarrow k = - 11$

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Question 5:

The sum of the roots of the equation $x^{2} - 6 x + 2 = 0$ is
(a) 2
(b) −2
(c) 6
(d) −6

Answer:

(c) 6
$Sum of the roots of the equation x^{2} - 6 x + 2 = 0 is α + β = \frac{- b}{a} = \frac{- (- 6)}{1} = 6 , where α and β are the roots of the equation.$

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Question 6:

If the product of the roots of the equation $x^{2} - 3 x + k = 10$ is −2, then the value of k is
(a) −2
(b) −8
(c) 8
(d) 12

Answer:

(c) 8
$It is given that the product of the roots of the equation x^{2} - 3 x + k = 10 is - 2 . The equation can be rewritten as : x^{2} - 3 x + (k - 10) = 0 Product of the roots of a quadratic equation = \frac{c}{a} \Rightarrow \frac{c}{a} = - 2 \Rightarrow \frac{(k - 10)}{1} = - 2 \Rightarrow k = 8$

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Question 7:

If one root of the equation $2 x^{2} + a x + 6 = 0$ is 2, then a = ?
(a) 7
(b) −7
(c) $\frac{7}{2}$
(d) $\frac{- 7}{2}$

Answer:

(b) −7
$It is given that one root of the equation 2 x^{2} + a x + 6 = 0 is 2. ∴ 2 \times 2^{2} + a \times 2 + 6 = 0 \Rightarrow 2 a + 14 = 0 \Rightarrow a = - 7$

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Question 8:

The ratio of the sum and product of the roots of the equation $7 x^{2} - 12 x + 18 = 0$
(a) 7 : 12
(b) 7 : 18
(c) 2 : 3
(d) 3 : 2

Answer:

(c) 2 : 3

$Given: 7 x^{2} - 12 x + 18 = 0 ∴ α + β = \frac{12}{7} and α β = \frac{18}{7}, where α and β are the roots of the equation ∴ Ratio of the sum and product of the roots = \frac{12}{7} : \frac{18}{7} = 12 : 18 = 2 : 3$

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Question 9:

The roots of the equation $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ are
(a) $\frac{2 \sqrt{3}}{3}, \frac{- \sqrt{3}}{4}$
(b) $\frac{- 2 \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$
(c) $\frac{\sqrt{3}}{3}, \frac{- \sqrt{3}}{4}$
(d) $\frac{- \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$

Answer:

(b) $\frac{- 2 \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$

$Given: 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0 \Rightarrow 4 \sqrt{3} x^{2} + 8 x - 3 x - 2 \sqrt{3} = 0 ⇒ 4 x (\sqrt{3} x + 2) - \sqrt{3} (\sqrt{3} x + 2) = 0 ⇒ (\sqrt{3} x + 2) (4 x - \sqrt{3}) = 0 ⇒ x = \frac{\sqrt{3}}{4} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{- 2 \sqrt{3}}{3} Thus, the roots of the equation are \frac{\sqrt{3}}{4} and \frac{- 2 \sqrt{3}}{3} .$

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Question 10:

The root of the equation $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ are
(a) $\sqrt{2}, \frac{5 \sqrt{2}}{2}$
(b) $- \sqrt{2}, \frac{5 \sqrt{2}}{2}$
(c) $\sqrt{2}, \frac{- 5 \sqrt{2}}{2}$
(d) $- \sqrt{2}, \frac{- 5 \sqrt{2}}{2}$

Answer:

(d) $- \sqrt{2}, \frac{- 5 \sqrt{2}}{2}$

$Given: \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 ⇒ x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 ⇒ (\sqrt{2} x + 5) (x + \sqrt{2}) = 0 ⇒ x = \frac{- 5}{\sqrt{2}} = \frac{- 5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{- 5 \sqrt{2}}{2} and x = - \sqrt{2}$

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Question 11:

The roots of the equation $3^{x + 2} + 3^{- x}$ = 10 are
(a) 2, 0
(b) −2, 0
(c) 3, −1
(d) −3, 1

Answer:

(b) −2, 0

$Given: 3^{x + 2} + 3^{- x} = 10 {Let 3}^{x} be equal to y . Then, the equation can be rewritten as: 3^{2} × 3^{x} + \frac{1}{3^{x}} = 10 \Rightarrow 9 y + \frac{1}{y} = 10 ⇒ \frac{9 y^{2} + 1}{y} = 10 ⇒ 9 y^{2} - 10 y + 1 = 0 ⇒ 9 y^{2} - 9 y - y + 1 = 0 \Rightarrow 9 y (y - 1) - (y - 1) = 0 ⇒ (y - 1) (9 y - 1) = 0 ⇒ y = 1 or y = \frac{1}{9} = \frac{1}{3^{2}} {⇒ 3}^{x} = 3^{0} {or 3}^{x} = 3^{- 2} ⇒ x = 0 or x = - 2$

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Question 12:

The roots of the equation $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ are
(a) $\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}$
(b) $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$
(c) $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

Answer:

(b) $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$

$Given: 3 x^{2} - 2 \sqrt{6} x + 2 = 0 We know that : x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{2 \sqrt{6} \pm \sqrt{{(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2}}{2 \times 3} = \frac{2 \sqrt{6} \pm \sqrt{24 - 24}}{6} = \frac{2 \sqrt{6} \pm 0}{6} ∴ x = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} = \frac{\sqrt{3} \times \sqrt{2}}{3} = \sqrt{\frac{2}{3}} Therefore, the roots are \sqrt{\frac{2}{3}} and \sqrt{\frac{2}{3}} .$

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Question 13:

The roots of the equation $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15}$ are
(a) $\frac{5}{2}, \frac{3}{2}$
(b) $\frac{5}{2}, \frac{- 3}{2}$
(c) $\frac{- 5}{2}, \frac{3}{2}$
(d) $\frac{- 5}{2}, \frac{- 3}{2}$

Answer:

(c) $\frac{- 5}{2}, \frac{3}{2}$

$Given : \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15} ⇒ \frac{x^{2} + {(x + 1)}^{2}}{(x + 1) x} = \frac{34}{15} ⇒ x^{2} + x^{2} + 2 x + 1 = \frac{34 (x^{2} + x)}{15} ⇒ 15 (2 x^{2} + 2 x + 1) = 34 x^{2} + 34 x \Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x \Rightarrow 4 x^{2} + 4 x - 15 = 0 ⇒ 4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0 \Rightarrow (2 x + 5) (2 x - 3) = 0 ⇒ x = \frac{- 5}{2} or x = \frac{3}{2} Thus, the roots are \frac{- 5}{2} and \frac{3}{2} .$

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Question 14:

The roots of the equation $\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{1 - x}{x}} = 2 \frac{1}{6}$ are
(a) $\frac{9}{13}, \frac{7}{13}$
(b) $\frac{9}{13}, \frac{4}{13}$
(c) $\frac{7}{13}, \frac{4}{13}$
(d) none of these

Answer:

(b) $\frac{9}{13}, \frac{4}{13}$

$Given: \sqrt{\frac{x}{1 - x}} + \sqrt{\frac{1 - x}{x}} = 2 \frac{1}{6} Let \frac{x}{(1 - x)} be equal to y^{2} . Then, the equation becomes: y + \frac{1}{y} = \frac{13}{6} ⇒ \frac{y^{2} + 1}{y} = \frac{13}{6} \Rightarrow 6 y^{2} - 13 y + 6 = 0 \Rightarrow 6 y^{2} - 9 y - 4 y + 6 = 0 \Rightarrow 3 y (2 y - 3) - 2 (2 y - 3) = 0 \Rightarrow (2 y - 3) (3 y - 2) = 0 \Rightarrow y = \frac{3}{2} or y = \frac{2}{3} ∴ \frac{x}{1 - x} = {(\frac{3}{2})}^{2} or \frac{x}{1 - x} = {(\frac{2}{3})}^{2} \Rightarrow \frac{x}{1 - x} = \frac{9}{4} or \frac{x}{1 - x} = \frac{4}{9} \Rightarrow 4 x = 9 - 9 x or 9 x = 4 - 4 x \Rightarrow 13 x = 9 or 13 x = 4 \Rightarrow x = \frac{9}{13} or x = \frac{4}{13} Therefore, the re quired roots are \frac{9}{13} and \frac{4}{13} .$

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Question 15:

The roots of $\frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = \frac{10}{3}$ are
(a) ±4
(b) ±6
(c) ±8
(d) $2 \pm \sqrt{3}$

Answer:

(c) ±8
$Given: \frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = \frac{10}{3} \Rightarrow \frac{{(x + 4)}^{2} + {(x - 4)}^{2}}{(x - 4) (x + 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} + 8 x + 16 + x^{2} - 8 x + 16}{x^{2} - 16} = \frac{10}{3} \Rightarrow 3 (2 x^{2} + 32) = 10 (x^{2} - 16) \Rightarrow 6 x^{2} + 96 = 10 x^{2} - 160 ⇒ 4 x^{2} = 256 ⇒ x^{2} = 64 \Rightarrow x = \pm 8$

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Question 16:

The root of a quadratic equation are 5 and −2. Then, the equation is
(a) $x^{2} - 3 x + 10 = 0$
(b) $x^{2} - 3 x - 10 = 0$
(c) $x^{2} + 3 x - 10 = 0$
(d) $x^{2} + 3 x + 10 = 0$

Answer:

(b) $x^{2} - 3 x - 10 = 0$

$It is given that the roots of the quadratic equation are 5 and - 2 . Then, the equation is: x^{2} - (5 - 2) x + 5 \times (- 2) = 0 ⇒ x^{2} - 3 x - 10 = 0$

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Question 17:

If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is
(a) $x^{2} - 6 x + 6 = 0$
(b) $x^{2} + 6 x - 6 = 0$
(c) $x^{2} - 6 x - 6 = 0$
(d) $x^{2} + 6 x + 6 = 0$

Answer:

(a) $x^{2} - 6 x + 6 = 0$

$Given : Sum of roots = 6 Product of roots = 6 Thus, the equation is: x^{2} - 6 x + 6 = 0$

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Question 18:

If one root of the equation $3 x^{2} - 10 x + 3 = 0 is \frac{1}{3},$ then the other root is
(a) $\frac{- 1}{3}$
(b) $\frac{1}{3}$
(c) −3
(d) 3

Answer:

(d) 3
$Given: 3 x^{2} - 10 x + 3 = 0 One root of the equation is \frac{1}{3} . Let the other root be α . We know that : Product of the roots = \frac{c}{a} \Rightarrow \frac{1}{3} \times α = \frac{3}{3} \Rightarrow α = 3$

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Question 19:

The quadratic equation whose one root is $(3 + 2 \sqrt{3})$ is
(a) $x^{2} + 6 x - 3 = 0$
(b) $x^{2} - 6 x - 3 = 0$
(c) $x^{2} + 6 x + 3 = 0$
(d) $x^{2} - 6 x + 3 = 0$

Answer:

(b) $x^{2} - 6 x - 3 = 0$

$It is given that one root of the equation is (3 + 2 \sqrt{3}) . Therefore, the other root will be (3 - 2 \sqrt{3}) . ∴ α + β = 6 and α β = (3 + 2 \sqrt{3}) (3 - 2 \sqrt{3}) = 9 - 12 = - 3 Therefore, the r equired equation is {x^{2} - 6 x + (- 3) = 0} . \Rightarrow x^{2} - 6 x - 3 = 0$

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Question 20:

If the sum of the roots of the equation $k x^{2} + 2 x + 3 x = 0$ is equal to their product, then the value of k is
(a) $\frac{1}{3}$
(b) $\frac{- 1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{- 2}{3}$

Answer:

(d) $\frac{- 2}{3}$
$Given: k x^{2} + 2 x + 3 k = 0 Sum of the roots = Product of the roots ⇒ \frac{- 2}{k} = \frac{3 k}{k} ⇒ 3 k = - 2 ⇒ k = \frac{- 2}{3}$

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Question 21:

If one root of $5 x^{2} + 13 x + k = 0$ be the reciprocal of the other root, then the value of k is
(a) 0
(b) 1
(c) 2
(d) 5

Answer:

(d) 5
$Let the roots of the equation (5 x^{2} + 13 x + k = 0) be α and \frac{1}{α} . ∴ Product of the roots = \frac{c}{a} \Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

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Question 22:

The roots of the equation $a x^{2} + b x + c = 0$ will be reciprocal of each other if
(a) a = b
(b) b= c
(c) c = a
(d) none of these

Answer:

(c) c = a
$Let the roots of the equation (a x^{2} + b x + c = 0) be α and \frac{1}{α} . ∴ Product of the roots = α \times \frac{1}{α} = 1 \Rightarrow \frac{c}{a} = 1 \Rightarrow c = a$

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Question 23:

If the roots of the equation $a x^{2} + b x + c = 0$ are equal, then then c = ?
(a) $\frac{- b}{2 a}$
(b) $\frac{b}{2 a}$
(c) $\frac{- b^{2}}{4 a}$
(d) $\frac{b^{2}}{4 a}$

Answer:

(d) $\frac{b^{2}}{4 a}$

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

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Question 24:

If the equation has equal roots, then k = ?
(a) 2 or 0
(b) −2 or 0
(c) 2 or −2
(d) 0 only

Answer:

(c) 2 or −2
$It is given that the roots of the equation (9 x^{2} + 6 k x + 4 = 0) are equal. ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(6 k)}^{2} - 4 \times 9 \times 4 = 0 \Rightarrow 36 k^{2} = 144 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$

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Question 25:

If the equation $x^{2} + 2 (k + 2) x + 9 k = 0$ has equal rots, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

Answer:

(a) 1 or 4

$It is given that the roots of the equation (x^{2} + 2 (k + 2) x + 9 k = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {2 (k + 2)}^{2} - 4 \times 1 \times 9 k = 0 \Rightarrow 4 (k^{2} + 4 k + 4) - 36 k = 0 \Rightarrow 4 k^{2} + 16 k + 16 - 36 k = 0 \Rightarrow 4 k^{2} - 20 k + 16 = 0 \Rightarrow k^{2} - 5 k + 4 = 0 \Rightarrow k^{2} - 4 k - k + 4 = 0 \Rightarrow k (k - 4) - (k - 4) = 0 \Rightarrow (k - 4) (k - 1) = 0 \Rightarrow k = 4 or k = 1$

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Question 26:

If the equation $4 x^{2} - 3 k x + 1 = 0$ has equal roots, then k = ?
(a) $\pm \frac{2}{3}$
(b) $\pm \frac{1}{3}$
(c) $\pm \frac{3}{4}$
(d) $\pm \frac{4}{3}$

Answer:

(d) $\pm \frac{4}{3}$

$It is given that the roots of the equation (4 x^{2} - 3 k x + 1 = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(3 k)}^{2} - 4 \times 4 \times 1 = 0 \Rightarrow 9 k^{2} = 16 \Rightarrow k^{2} = \frac{16}{9} \Rightarrow k = \pm \frac{4}{3}$

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Question 27:

If the equation $x^{2} - 2 x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots, then k = ?
(a) $2 or \frac{10}{9}$
(b) $- 2 or \frac{10}{9}$
(c) $2 or \frac{- 10}{9}$
(d) $- 2 or \frac{- 10}{9}$

Answer:

(c) $2 or \frac{- 10}{9}$

$It is given that the roots of the equation (x^{2} - 2 x (1 + 3 k) + 7 (3 + 2 k) = 0 are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {2 (1 + 3 k)}^{2} - 4 \times 1 \times 7 (3 + 2 k) = 0 \Rightarrow 4 (9 k^{2} + 6 k + 1) - 28 (3 + 2 k) = 0 \Rightarrow 36 k^{2} + 24 k + 4 - 84 - 56 k = 0 \Rightarrow 36 k^{2} - 32 k - 80 = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0 \Rightarrow (k - 2) (9 k + 10) = 0 \Rightarrow k = 2 or k = \frac{- 10}{9}$

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Question 28:

If the equation $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$ has both roots equal, then
(a) b = ac
(b) $b = \frac{1}{2} (a + c)$
(c) b² = ac
(d) $b = \frac{2 a c}{(a + c)}$

Answer:

(c) b² = ac

$It is given that the roots of the equation {(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0} are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {2 b (a + c)}^{2} - 4 (a^{2} + b^{2}) (b^{2} + c^{2}) = 0 \Rightarrow 4 b^{2} (a^{2} + 2 a c + c^{2}) - 4 (a^{2} b^{2} + a^{2} c^{2} + b^{4} + b^{2} c^{2}) = 0 \Rightarrow 4 a^{2} b^{2} + 8 a b^{2} c + 4 b^{2} c^{2} - 4 a^{2} b^{2} - 4 a^{2} c^{2} - 4 b^{4} - 4 b^{2} c^{2} = 0 \Rightarrow 8 a b^{2} c - 4 a^{2} c^{2} - 4 b^{4} = 0 \Rightarrow 2 a b^{2} c - a^{2} c^{2} - b^{4} = 0 \Rightarrow b^{4} - 2 a c b^{2} + a^{2} c^{2} = 0 \Rightarrow {(b^{2} - a c)}^{2} = 0 \Rightarrow b^{2} - a c = 0 \Rightarrow b^{2} = a c$

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Question 29:

The roots of $a x^{2} + b x + c = 0, a \neq 0$ are real and unequal, if $(b^{2} - 4 a c)$
(a) > 0
(b) = 0
(c) < 0
(d) none of these

Answer:

(a) > 0
$\begin{array}{l} The roots of the equation are real and unequal when (b^{2} - 4 a c) > 0 . \end{array}$

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Question 30:

In the equation $a x^{2} + b x + c = 0,$ it is given that $D = (b^{2} - 4 a c) > 0 .$ Then, the roots of the equation are
(a) real and equal
(b) real and unequal
(c) imaginary
(d) none of these

Answer:

(b) real and unequal
$If D = (b^{2} - 4 a c) > 0, the roots are real and unequal .$

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Question 31:

If the equation $x^{2} + 5 k x + 16 = 0$ has no real roots, then
(a) $k > \frac{8}{5}$
(b) $k < \frac{- 8}{5}$
(c) $\frac{- 8}{5} < k > \frac{8}{5}$
(d) none of these

Answer:

(c) $\frac{- 8}{5} < k > \frac{8}{5}$

$It is given that the equation (x^{2} + 5 k x + 16 = 0) has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(5 k)}^{2} - 4 \times 1 \times 16 < 0 \Rightarrow 25 k^{2} - 64 < 0 \Rightarrow k^{2} < \frac{64}{25} \Rightarrow \frac{- 8}{5} < k < \frac{8}{5}$

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Question 32:

If the equation $x^{2} - k x + 1 = 0$ has no real roots, then
(a) k < −2
(b) k > 2
(c) −2 < k < 2
(d) none of these

Answer:

(c) −2 < k < 2

$It is given that the equation x^{2} - k x + 1 = 0 has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(- k)}^{2} - 4 \times 1 \times 1 < 0 \Rightarrow k^{2} < 4 \Rightarrow - 2 < k < 2$

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Question 33:

The roots of the equation $2 x^{2} - 6 x + 7 = 0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(d) imaginary

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 7 = 36 - 56 = - 20 < 0 Thus, the roots of the equation are imaginary .$

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Question 34:

The roots of the equation $2 x^{2} - 6 x + 3 = 0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(b) real, unequal and irrational

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 3 = 36 - 24 = 12 12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational .$

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Question 35:

For the equation $a x^{2} + b x + c = 0$ which of the following statements is incorrect?
(a) If (b² − 4ac) < 0, the roots are imaginary.
(b) If (b² − 4ac) = 0, the roots are real and equal.
(c) If (b² − 4ac) > 0 and (b² − 4ac) is a perfect square, then the roots are rational and unequal.
(d) (b² − 4ac) < 0, the roots are irrational.

Answer:

(d) (b²− 4ac) < 0, the roots are irrational.

$If (b^{2} - a c) < 0, the roots may be rational or irrational .$

Page No 465:

Question 36:

If the roots of $5 x^{2} - k x + 1 = 0$ are real and distinct, then
(a) $- 2 \sqrt{5} < k < 2 \sqrt{5}$
(b) $k > 2 \sqrt{5}$ only
(c) $k < - 2 \sqrt{5}$ only
(d) either $k > 2 \sqrt{5}$ or $k < - 2 \sqrt{5}$

Answer:

(d) either $k > 2 \sqrt{5}$ or $k < - 2 \sqrt{5}$

$It is given that the roots of the equation (5 x^{2} - k x + 1 = 0) are real and distinct . ∴ (b^{2} - 4 a c) > 0 \Rightarrow {(- k)}^{2} - 4 \times 5 \times 1 > 0 \Rightarrow k^{2} - 20 > 0 \Rightarrow k^{2} > 20 \Rightarrow k > \sqrt{20} or k < - \sqrt{20} \Rightarrow k > 2 \sqrt{5} or k < - 2 \sqrt{5}$

Page No 466:

Question 37:

The root of the equation $3 x^{2} + 7 x + 8 = 0$ are
(a) both real and equal
(b) both real and unequal
(c) both imaginary
(d) none of these

Answer:

(c) both imaginary

$∵ D = (b^{2} - 4 a c) = 49 - 4 \times 3 \times 8 = 49 - 96 = - 47 < 0 Thus, the roots of the equation are imaginary .$

Page No 466:

Question 38:

The sum of a number and its reciprocal is $2 \frac{1}{20} .$ The number is
(a) $\frac{5}{4}$
(b) $\frac{4}{3}$
(c) $\frac{3}{4}$
(d) $\frac{1}{6}$

Answer:

(a) $\frac{5}{4}$

$Let the required number be x . According to the question : x + \frac{1}{x} = \frac{41}{20} \Rightarrow \frac{x^{2} + 1}{x} = \frac{41}{20} \Rightarrow 20 x^{2} - 41 x + 20 = 0 \Rightarrow 20 x^{2} - 25 x - 16 x + 20 = 0 \Rightarrow 5 x (4 x - 5) - 4 (4 x - 5) = 0 \Rightarrow (4 x - 5) (5 x - 4) = 0 \Rightarrow x = \frac{5}{4} or x = \frac{4}{5}$

Page No 466:

Question 39:

The two parts into which 57 should be divided so that their product is 782, are
(a) 43 and 14
(b) 33 and 24
(c) 34 and 23
(d) 44 and 13

Answer:

(c) 34 and 23

$Let the required numbers be x and (57 - x) . According to the question : x (57 - x) = 782 \Rightarrow 57 x - x^{2} = 782 \Rightarrow x^{2} - 57 x + 782 = 0 \Rightarrow x^{2} - 34 x - 23 x + 782 = 0 \Rightarrow x (x - 34) - 23 (x - 34) = 0 \Rightarrow (x - 34) (x - 23) = 0 \Rightarrow x = 34 or x = 23 T hus, the required numbers are 34 and 23.$

Page No 466:

Question 40:

The perimeter of a rectangle is 82 m and its area is 400 m². The breadth of the rectangle is
(a) 25 m
(b) 20 m
(c) 16 m
(d) 9 m

Answer:

(c) 16 m

$Let the length and breadth of the rectangle be l and b . Perimeter of the rectangle = 82 m \Rightarrow 2 \times (l + b) = 82 \Rightarrow l + b = 41 \Rightarrow l = (41 - b) . . . (i) Area of the rectangle = 400 m^{2} \Rightarrow l \times b = 400 m^{2} \Rightarrow (41 - b) b = 400 (using (i)) \Rightarrow 41 b - b^{2} = 400 \Rightarrow b^{2} - 41 b + 400 = 0 \Rightarrow b^{2} - 25 b - 16 b + 400 = 0 \Rightarrow b (b - 25) - 16 (b - 25) = 0 \Rightarrow (b - 25) (b - 16) = 0 \Rightarrow b = 25 or b = 16 If b = 25, we have : l = 41 - 25 = 16 Since, l cannot be less than b, ∴ b = 16 m$

Page No 466:

Question 41:

Which constant should be added and subtracted to solve the quadratic equation $4 x^{2} - \sqrt{3} x - 5 = 0$ by the method of completing the square?
(a) $\frac{9}{10}$
(b) $\frac{3}{16}$
(c) $\frac{3}{4}$
(d) $\frac{\sqrt{3}}{4}$

Answer:

(b) $\frac{3}{16}$

$Given : 4 x^{2} - \sqrt{3} x - 5 = 0 \Rightarrow {(2 x)}^{2} - 2 \times 2 x \times \frac{\sqrt{3}}{4} + {(\frac{\sqrt{3}}{4})}^{2} - {(\frac{\sqrt{3}}{4})}^{2} - 5 = 0 \Rightarrow {(2 x - \frac{\sqrt{3}}{4})}^{2} + \frac{3}{16} - 5 = 0 Clearly, if \frac{3}{16} is added to the equation, the equation becomes a perfect square .$

Page No 466:

Question 42:

The roots of the equation $2 x - \frac{3}{x} = 1$ are
(a) $\frac{1}{2}, - 1$
(b) $\frac{3}{2}, 1$
(c) $\frac{3}{2}, - 1$
(d) $\frac{- 1}{2}, \frac{3}{2}$

Answer:

(c) $\frac{3}{2}, - 1$

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

Page No 466:

Question 43:

For what real values of k, the equation $9 x^{2} + 8 k x + 16 = 0$ has real and equal roots?
(a) k = 2 or −2
(b) k = 3 or −3
(c) $k = \frac{4}{3} or \frac{- 4}{3}$
(d) None of these

Answer:

(b) k = 3 or −3

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

Page No 466:

Question 44:

For what values of k, the equation $k x^{2} - 6 x - 2 = 0$ has real roots?
(a) $k \leq \frac{- 9}{2}$
(b) $k \geq \frac{- 9}{2}$
(c) $k \leq - 2$
(d) None of these

Answer:

(b) $k \geq \frac{- 9}{2}$

$It is given that the roots of the equation (k x^{2} - 6 x - 2 = 0) are real. ∴ D \geq 0 \Rightarrow (b^{2} - 4 a c) \geq 0 \Rightarrow {(- 6)}^{2} - 4 \times k \times (- 2) \geq 0 \Rightarrow 36 + 8 k \geq 0 \Rightarrow k \geq \frac{- 36}{8} \Rightarrow k \geq \frac{- 9}{2}$

Page No 466:

Question 45:

If α and β are the roots of the equation $3 x^{2} + 8 x + 2 = 0, then (\frac{1}{α} + \frac{1}{β}) = ?$
(a) $\frac{- 3}{8}$
(b) $\frac{2}{3}$
(c) −4
(d) 4

Answer:

(c) −4

$Given: α and β are the roots of the equation 3 x^{2} + 8 x + 2 = 0 ∴ α + β = \frac{- 8}{3} and α β = \frac{2}{3} ...(i) ∴ (\frac{1}{α} + \frac{1}{β}) = (\frac{α + β}{α β}) = \frac{(\frac{- 8}{3})}{(\frac{2}{3})} (from (i)) \begin{array}{l} = \frac{- 8}{2} \end{array} = - 4$

Page No 466:

Question 46:

Which of the following is not true?
(a) Every quadratic equation can have at the most two real roots.
(b) Some quadratic equations do not have any real root.
(c) Some quadratic equations may have one real root.
(d) Every quadratic equations has two real roots.

Answer:

$(d) Every quadratic equations has two real roots. An equation may or may not have real roots .$

Page No 467:

Question 47:

Consider the following statements:
I. If the roots of the equation $a x^{2} + b x + c = 0$ are negative reciprocal of each other, than a + c = 0.
II. A quadratic equation can have at all most two roots.
III. If α, β are the roots of $x^{2} - 22 x + 105 = 0,$ then α + β = 22 and α − β = 8.

Of these statements:
(a) I and II are true and III is false.
(b) I and III are true and II is false.
(c) II and III are true I is false.
(d) I, II and III are all true.

Answer:

(d) I, II and III are all true.

$(I) Let the roots be a and - \frac{1}{a} . Then, a \times (\frac{- 1}{a}) = \frac{c}{a} \begin{array}{l} \Rightarrow c = - a \\ \Rightarrow a + c = 0 \end{array} ∴ I is true . (II) Clearly, II is true . (III) α + β = 22 and α β = (15 \times 7) = 105 On solving them, we get : α = 15 and β = 7 α - β = 15 - 7 = 8 ∴ III is true . ∴ I, II and III are all true .$

Page No 467:

Question 48:

Assertion (A)
The equation x² + x + 1 = 0 has both real roots.

Reason (R)
The equation ax2 + bx + c = 0, (a ≠ 0) has both real roots, if (b2 − 4ac) > 0.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false and Reason (R) is true.

$Reason (R) is true . For the equation x^{2} + x + 1 = 0, we have: D = 1^{2} - 4 \times 1 \times 1 = - 3 < 0 This implies that b oth the roots of the equation are imaginary . Therefore, Assertion (A) is false . Thus, Reason (R) is true and Assertion (A) is false .$

Page No 467:

Question 49:

Assertion (A)
The equation $2 x^{2} - 4 x + 3 = 0$ has no real roots.

Reason (R)
The equation $a x^{2} + b x + c = 0, (a \neq 0)$ has no real roots, if $(b^{2} - 4 a c) < 0 .$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

$Reason (R) is true . F rom the equation 2 x^{2} - 4 x + 3 = 0, we have: D = {(- 4)}^{2} - 4 \times 2 \times 3 = 16 - 24 = - 8 < 0 Therefore, the given equation has no real roots . Thus, Assertion (A) is true and Reason (R) is a correct explanation of Assertion (A) .$

Page No 467:

Question 50:

Assertion (A)
If −5 is a root of $2 x^{2} + 2 p x - 15 = 0 and p (x^{2} + x) + k = 0$ has equal roots, then $k = \frac{7}{8} .$

Reason (R)
The equation $a x^{2} + b x + c = 0, a \neq 0$ has equal roots, if $(b^{2} - 4 a c) = 0 .$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

$Reason (R) is true . It is given that - 5 is a root of 2 x^{2} + 2 p x - 15 = 0; therefore, we have: 2 \times {(- 5)}^{2} + 2 p \times (- 5) - 15 = 0 \Rightarrow 50 - 10 p - 15 = 0 \Rightarrow 10 p = 35 \Rightarrow p = \frac{35}{10} = \frac{7}{2} Now, \frac{7}{2} (x^{2} + x) + k = 0 \Rightarrow 7 x^{2} + 7 x + 2 k = 0 For equal roots, we have: D = 0 \Rightarrow 49 - 4 \times 7 \times 2 k = 0 \Rightarrow 56 k = 49 \Rightarrow k = \frac{49}{56} = \frac{7}{8} Thus, Assertion (A) is true and Reason (R) is a correct explanation of Assertion (A) .$

Page No 468:

Question 51:

Assertion (A)
The roots of $2 x^{2} + x - 6 = 0 are - 2 and \frac{3}{2} .$

Reason (R)
If $a x^{2} + b x + c, a \neq 0, t h e n x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

$Reason (R) is true . \begin{array}{l} For the equation 2 x^{2} + x - 6 = 0, we have : \end{array}$
$x = \frac{- 1 \pm \sqrt{1^{2} - 4 \times 2 \times (- 6)}}{2 \times 2} \Rightarrow x = \frac{- 1 \pm \sqrt{1 + 48}}{4} \Rightarrow x = \frac{- 1 \pm \sqrt{49}}{4} \Rightarrow x = \frac{- 1 \pm 7}{4} \Rightarrow x = \frac{6}{4}, \frac{- 8}{4} \Rightarrow x = \frac{3}{2}, - 2$

Thus, Assertion (A) is true and Reason (R) is a correct explanation of Assertion (A).

Page No 468:

Question 52:

Assertion (A)
If $p x^{2} - 2 x + 2 = 0$ has real roots, then $p \leq \frac{1}{2} .$

Reason (R)
The equation $(a^{2} +^{b}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ has no real root, if ad ≠ bc.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

$For no real roots, we have: (b^{2} - 4 a c) < 0 Now, 4 {(a c + b d)}^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2}) = 4 a^{2} c^{2} + 4 b^{2} d^{2} + 8 a b c d - 4 a^{2} c^{2} - 4 a^{2} d^{2} - 4 b^{2} c^{2} - 4 b^{2} d^{2} = - 4 (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}) = - 4 {(a d - b c)}^{2} < 0, when (a d - b c) \neq 0 \Rightarrow a d \neq b c Therefore, Reason (R) is true . Now, p x^{2} - 2 x + 2 = 0 has real r o o t s; therefore, we have : (b^{2} - 4 a c) \geq 0 \Rightarrow 4 - 4 \times p \times 2 \geq 0 \Rightarrow 4 - 8 p \geq 0 \Rightarrow 8 p \leq 4 \Rightarrow p \leq \frac{1}{2} Therefore, Assertion (A) is true . Thus, both Reason (R) and Assertion (A) are true but Reason (R) is not a correct explanation of Assertion (A) .$

Page No 468:

Question 53:

Match the following columns:

Column I	Column II
(a) $2 x^{2} - 7 x + 5 = 0$	(p) Real and equal
(b) $3 x^{2} - 2 \sqrt{6} x + 2 = 0$	(q) Real, distinct and irrational
(c) $9 x^{2} - 10 x + 15 = 0$	(r) Real distinct and rational
(d) $\sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0$	(s) Imaginary

Answer:

$\begin{array}{l} (a) 2 x^{2} - 7 x + 5 = 0 \\ D = b^{2} - 4 a c = 49 - 40 = 9 > 0 Also, D is not a perfect square . Therefore, the roots are real, distinct and rational . (b) 3 x^{2} - 2 \sqrt{6} x + 2 = 0 \begin{array}{l} D = b^{2} - 4 a c = 24 - 24 = 0 \\ Therefore, the roots are real and equal . \end{array} \begin{array}{l} (c) 9 x^{2} - 10 x + 15 = 0 \\ D = b^{2} - 4 a c = 100 - 540 = - 440 < 0 \end{array} Therefore, t he roots are imaginary . \begin{array}{l} (d) \sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0 \\ D = b^{2} - 4 a c = 8 + 24 = 32 > 0 \end{array} \begin{array}{l} Also, D is not a perfect square . Therefore, t he roots are real, distinct and irrational . \\ So, the correct options are (a) - (r), (b) - (p), (c) - (s) and (d) - (q) . \end{array} \end{array}$

Page No 475:

Question 1:

If $x^{2} - 4 x + p = 0$ has real roots, then
(a) p ≥ 4
(b) p ≤ 4
(c) p ≥ 5
(d) p ≤ 5

Answer:

(b) p ≤ 4
$It is given that the equation (x^{2} + 4 x + p = 0) has real roots. ∴ D \geq 0 \Rightarrow b^{2} - 4 a c \geq 0 \Rightarrow 16 - 4 p \geq 0 \Rightarrow 16 \geq 4 p \Rightarrow p \leq 4$

Page No 475:

Question 2:

If x = 1 is a common root of $a x^{2} + a x + 3 = 0 and x^{2} + x + b = 0, then a b = ?$
(a) 3
(b) −3
(c) 4
(d) 6

Answer:

(a) 3

$It is given that x = 1 is a common root of (a x^{2} + a x + 3 = 0) and (x^{2} + x + b = 0) . ∴ a {(1)}^{2} + a (1) + 3 = 0 and {(1)}^{2} + 1 + b = 0 \Rightarrow a = \frac{- 3}{2} and b = - 2 ∴ a b = \frac{- 3}{2} \times (- 2) = 3$

Page No 475:

Question 3:

If $x^{2} + 2 (k + 2) x + 9 k = 0$ has a repeated root, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

Answer:

(a) 1 or 4

$It is given that the equation (x^{2} + 2 (k + 2) x + 9 k = 0) has repeated roots, i .e ., roots are equal . ∴ D = 0 \Rightarrow 4 {(k + 2)}^{2} - 4 \times 1 \times 9 k = 0 \Rightarrow 4 k^{2} + 16 k + 16 - 36 k = 0 \Rightarrow 4 k^{2} - 20 k + 16 = 0 \Rightarrow k^{2} - 5 k + 4 = 0 \Rightarrow k^{2} - 4 k - k + 4 = 0 \Rightarrow k (k - 4) - 1 (k - 4) = 0 \Rightarrow (k - 1) (k - 4) = 0 \Rightarrow k = 1 or 4$

Page No 475:

Question 4:

If $x^{2} + 5 p x + 16 = 0$ has no real root, then
(a) $p > \frac{8}{5}$
(b) $p < \frac{- 8}{5}$
(c) $\frac{- 8}{5} < p < \frac{8}{5}$
(d) none of these

Answer:

(c) $\frac{- 8}{5} < p < \frac{8}{5}$

$It is given that the equation (x^{2} + 5 p x + 16 = 0) has no real roots. ∴ D < 0 \Rightarrow b^{2} - 4 a c < 0 \Rightarrow 25 p^{2} - 4 \times 16 < 0 \Rightarrow 25 p^{2} < 64 \Rightarrow p^{2} < \frac{64}{25} \Rightarrow \frac{- 8}{5} < p < \frac{8}{5}$

Page No 476:

Question 5:

For what values of k, the roots the equation $x^{2} - k x + 1 = 0$ are imaginary?

Answer:

$\begin{array}{l} It is given that the roots of the equation (x^{2} - k x + 1 = 0) are imaginary. \\ Here, a = 1, b = - k and c = 1 \end{array} ∴ b^{2} - 4 a c < 0 \Rightarrow k^{2} - 4 \times 1 \times 1 < 0 \Rightarrow k^{2} < 4 \Rightarrow k^{2} - 4 < 0 \Rightarrow (k + 2) (k - 2) < 0 \Rightarrow - 2 < k < 2$

Page No 476:

Question 6:

For what values of k, the roots of the equation $x^{2} + 4 k + k = 0$ are real?

Answer:

$It is given that the roots of the equation (x^{2} + 4 x + k = 0) are real. ∴ b^{2} - 4 a c \geq 0 Here, a = 1, b = - 4 and c = k ⇒ 16 - 4 \times 1 \times k \geq 0 ⇒ 16 \geq 4 k ∴ k \leq 4$
For real roots, the value of k must be lesser than or equal to 4.

Page No 476:

Question 7:

For what values of k, the roots of the equation $2 x^{2} + k x + k = 0$ are real?

Answer:

$It is given that the roots of the equation (2 x^{2} - k x + k = 0) are equal. Here, a = 2, b = - k and c = k ∴ b^{2} - 4 a c = 0 ⇒ k^{2} - 4 \times 2 \times k = 0 ⇒ k^{2} = 8 k ⇒ k^{2} - 8 k = 0 ⇒ k (k - 8) = 0 ⇒ k = 0 or k = 8$

Page No 476:

Question 8:

Solve: $2 x^{2} - 6 x + 3 = 0 .$

Answer:

$Given: 2 x^{2} - 6 x + 3 = 0 By quadratic formula, we can find the roots of the equation . Therefore, we have : x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} Here, b = - 6, a = 2 and c = 3 \Rightarrow x = \frac{- (- 6) \pm \sqrt{(- 6)^{2} - 4 \times 2 \times 3}}{2 \times 2} \Rightarrow x = \frac{6 \pm \sqrt{36 - 24}}{4} \Rightarrow x = \frac{3 \pm \sqrt{3}}{2} \Rightarrow x = \frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}$

Page No 476:

Question 9:

Show that the equation $2 x^{2} - 5 x - 4 = 0$ has real and unequal roots.

Answer:

$We have to show that the roots of the equation (2 x^{2} - 5 x - 4 = 0) are real and unequal . For real and unequal roots, we have : D > 0 o r b^{2} - 4 a c > 0 Here, b = - 5, a = 2 and c = - 4 ∴ b^{2} - 4 a c = (- 5)^{2} - 4 \times 2 \times (- 4) = 25 + 32 = 57 > 0 For the given equation, D > 0; therefore, the roots are real and unequal .$

Page No 476:

Question 10:

Show that the equation $x^{2} - x + 2 = 0$ has no real roots.

Answer:

$x^{2} - x + 2 = 0 H ere, a = 1, b = - 1 and c = 2 ∴ b^{2} - 4 a c = 1 - 4 \times 1 \times 2 = - 7 < 0 D < 0 for the equation; therefore, it has no real solution.$

Page No 476:

Question 11:

Solve: $10 x - \frac{1}{x} = 3 .$

Answer:

$10 x - \frac{1}{x} = 3 \Rightarrow 10 x^{2} - 1 = 3 x ⇒ 10 x^{2} - 3 x - 1 = 0 ⇒ 10 x^{2} - 5 x + 2 x - 1 = 0 ⇒ 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0 ∴ x = \frac{1}{2}, x = - \frac{1}{5}$

Page No 476:

Question 12:

Solve: $\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0$

Answer:

$\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0 ⇒ \sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0 ⇒ \sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0 ⇒ (x + 3 \sqrt{3}) (\sqrt{3} x + 2) = 0 ∴ x = - 3 \sqrt{3}, x = - \frac{2}{\sqrt{3}}$

Page No 476:

Question 13:

Solve: $3 x^{2} + 2 \sqrt{5} x - 5 = 0 .$

Answer:

$3 x^{2} + 2 \sqrt{5} x - 5 = 0 ⇒ 3 x^{2} + 3 \sqrt{5} x - \sqrt{5} x - 5 = 0 ⇒ 3 x (x + \sqrt{5}) - \sqrt{5} (x + \sqrt{5}) = 0 ⇒ (x + \sqrt{5}) (3 x - \sqrt{5}) = 0 ∴ x = - \sqrt{5}, x = \frac{\sqrt{5}}{3}$

Page No 476:

Question 14:

Find two consecutive positive integers, the sum of whose squares is 25.

Answer:

$Let the two consecutive positive integers be x and (x + 1) . According to the question : x^{2} + {(x + 1)}^{2} = 25 ⇒ x^{2} + x^{2} + 2 x + 1 - 25 = 0 ⇒ 2 x^{2} + 2 x - 24 = 0 ⇒ x^{2} + x - 12 = 0 ⇒ x^{2} + (4 - 3) x - 12 = 0 ⇒ x^{2} + 4 x - 3 x - 12 = 0 ⇒ x (x + 4) - 3 (x + 4) = 0 \Rightarrow (x + 4) (x - 3) = 0 ∴ x = - 4, x = 3 It is given that the integers are positive; therefore the value of x will be 3 . Thus, the two positive integers are 3 and {(3 + 1) = 4}.$

Page No 476:

Question 15:

Solve: $x^{2} - 4 a x + 4 a^{2} - b^{2} = 0 .$

Answer:

$x^{2} - 4 a x + 4 a^{2} - b^{2} = 0 Here, a = 1, b = - 4 a and c = 4 a^{2} - b^{2} ∴ D = \sqrt{b^{2} - 4 a c} = \sqrt{16 a^{2} - 4 (4 a^{2} - b^{2})} = \sqrt{16 a^{2} - 16 a^{2} + 4 b^{2}} = 2 b ∴ The roots of the equation are: \frac{- b + D}{2 a}, \frac{- b - D}{2 a} or, \frac{4 a + 2 b}{2}, \frac{4 a - 2 b}{2} or, \frac{2 (2 a + b)}{2}, \frac{2 (2 a - b)}{2} or, (2 a + b), (2 a - b)$

Page No 476:

Question 16:

Using quadratic formula, solve $a b x^{2} + (b^{2} - a c) x - b c = 0 .$

Answer:

$a b x^{2} + (b^{2} - a c) x - b c = 0 Here, a = a b, b = b^{2} - a c and c = - b c D = \sqrt{b^{2} - 4 a c} = \sqrt{{(b^{2} - a c)}^{2} - 4 \times a b \times (- b c)} = \sqrt{b^{4} - 2 b^{2} a c + a^{2} c^{2} + 4 b^{2} a c} = \sqrt{b^{4} + 2 b^{2} a c + a^{2} c^{2}} = \sqrt{{(b^{2} + a c)}^{2}} = (b^{2} + a c) ∴ The roots of the equation are: \frac{- b + D}{2 a}, \frac{- b - D}{2 a} or, \frac{- (b^{2} - a c) + (b^{2} + a c)}{2 a b}, \frac{- (b^{2} - a c) - (b^{2} + a c)}{2 a b} or, \frac{- b^{2} + a c + b^{2} + a c}{2 a b}, \frac{- b^{2} + a c - b^{2} - a c}{2 a b} or, \frac{2 a c}{2 a b}, \frac{- 2 b^{2}}{2 a b} or, \frac{c}{b}, \frac{- b}{a}$

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Question 17:

The area of a right-angle triangle is 165 m². Determine its base and altitude if the latter exceeds the former by 7 metres.

Answer:

$Let the base of the right-angled triangle be x cm. According to the question : Altitude = (x + 7) cm ∴ A rea of the triangle = \frac{1}{2} \times x \times (x + 7) {cm}^{2} Now, \frac{1}{2} \times x \times (x + 7) = 165 ⇒ x^{2} + 7 x - 330 = 0 ⇒ x^{2} + (22 - 15) x - 330 = 0 ⇒ x^{2} + 22 x - 15 x - 330 = 0 ⇒ x (x + 22) - 15 (x + 22) = 0 \Rightarrow (x + 22) (x - 15) = 0 ∴ x = - 22, x = 15 The base of the triangle can never be negative. Therefore, the base of the right-angled triangle is 15 cm. Altitude = (15 + 7) = 22 cm$

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Question 18:

Solve: $\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3} .$

Answer:

$\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3} ⇒ (x + 3) (2 x - 3) = (x + 2) (3 x - 7) ⇒ 2 x^{2} - 3 x + 6 x - 9 = 3 x^{2} - 7 x + 6 x - 14 ⇒ 3 x^{2} - 7 x + 6 x - 14 - 2 x^{2} + 3 x - 6 x + 9 = 0 ⇒ x^{2} - 4 x - 5 = 0 ⇒ x^{2} - 5 x + x - 5 = 0 ⇒ x (x - 5) + 1 (x - 5) = 0 \Rightarrow (x - 5) (x + 1) = 0 ∴ x = 5, x = - 1$

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Question 19:

Solve: $2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5, x \neq - 3, \frac{1}{2} .$

Answer:

$2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5 ⇒ \frac{4 x - 2}{x + 3} - \frac{3 x + 9}{2 x - 1} = 5 ⇒ \frac{(4 x - 2) (2 x - 1) - (x + 3) (3 x + 9)}{(x + 3) (2 x - 1)} = 5 ⇒ 8 x^{2} - 4 x - 4 x + 2 - (3 x^{2} + 9 x + 9 x + 27) = 5 (2 x^{2} - x + 6 x - 3) ⇒ 8 x^{2} - 8 x + 2 - 3 x^{2} - 9 x - 9 x - 27 = 10 x^{2} - 5 x + 30 x - 15 ⇒ 10 x^{2} - 5 x^{2} + 25 x + 26 x - 15 + 25 = 0 ⇒ 5 x^{2} + 51 x + 10 = 0 \Rightarrow 5 x^{2} + 50 x + x + 10 = 0 \Rightarrow 5 x (x + 10) + 1 (x + 10) = 0 \Rightarrow (x + 10) (5 x + 1) = 0 ∴ x = - 10, x = - \frac{1}{5}$

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Question 20:

Solve: 5^(x+1)+5^(2−x)=126.

Answer:

$5^{x + 1} + 5^{2 - x} = 126 {⇒ 5}^{x} \times 5 + \frac{25}{5^{x}} = 126 ⇒ 5 y + \frac{25}{y} = 126 [where y = 5^{x}] ⇒ 5 y^{2} - 126 y + 25 = 0 ⇒ 5 y^{2} - 125 y - y + 25 = 0 ⇒ 5 y (y - 25) - 1 (y - 25) = 0 ⇒ (y - 25) (5 y - 1) = 0 ∴ y = 25, y = \frac{1}{5} or 5^{x} = 5^{2}, 5^{x} = 5^{- 1} ∴ x = 2, x = - 1$

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