Mathematics Part i Solutions Solutions for Class 10 Math Chapter 2 - Quadratic Equations

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Page No 34:

Question 1:

Write any two quadratic equations.

Answer:

Two quadratic equations are
$x^{2} + 10 x - 200 = 0$ and $x^{2} + 5 x - 6 = 0$ .

Page No 34:

Question 2:

Decide which of the following are quadratic equations.
(1) x² + 5x – 2 = 0
(2) y²= 5y – 10
(3) $y^{2} + \frac{1}{y} = 2$
(4) $x + \frac{1}{x} = - 2$
(5) (m + 2) (m – 5) = 0
(6) m³+ 3m² – 2 = 3m³

Answer:

(1) x² + 5x – 2 = 0
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(2) y²= 5y – 10
Only one variable y.
Maximum index = 2
So, it is a quadratic equation.

(3) $y^{2} + \frac{1}{y} = 2$
$\Rightarrow y^{3} + 1 = 2 y$
Only one variable y.
Maximum index = 3
So, it is not a quadratic equation.

(4) $x + \frac{1}{x} = - 2$
$\Rightarrow x^{2} + 1 = - 2 x$
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(5) (m + 2) (m – 5) = 0
$\Rightarrow m^{2} - 3 m - 10 = 0$
Only one variable m.
Maximum index = 2
So, it is a quadratic equation.

(6) m³+ 3m² – 2 = 3m³
Only one variable m.
Maximum index = 3
So, it is not a quadratic equation.

Page No 34:

Question 3:

Write the following equations in the form ax² + bx + c= 0, then write the values of a, b, c for each equation.
(1) 2y = 10 – y²
(2) (x – 1)² = 2x + 3
(3) x² + 5x = –(3 – x)
(4) 3m² = 2m² – 9
(5) P(3+6p) = –5
(6) x² – 9 = 13

Answer:

(1) 2y = 10 – y^{2
$\Rightarrow y^{2} + 2 y = 10 \Rightarrow y^{2} + 2 y - 10 = 0$}
So,it is of the form ax² + bx + c = 0 where a = 1, b = 2 and c = −10.

(2) (x – 1)² = 2x + 3
$\Rightarrow (x^{2} - 2 x + 1) = 2 x + 3 \Rightarrow x^{2} - 4 x - 2 = 0$
So,it is of the form ax² + bx + c = 0 where a = 1, b = −4 and c = −2.

(3) x² + 5x = –(3 – x)
$\Rightarrow x^{2} + 5 x + 3 - x = 0 \Rightarrow x^{2} + 4 x + 3 = 0$
So,it is of the form ax² + bx + c = 0 where a = 1, b = 4 and c = 3.

(4) 3m² = 2m² – 9
$\Rightarrow 3 m^{2} - 2 m^{2} + 9 = 0 \Rightarrow m^{2} + 0 m + 9 = 0$
So,it is of the form ax² + bx + c = 0 where a = 1, b = 0 and c = 9.

(5) p(3+6p) = –5
$\Rightarrow 3 p + 6 p^{2} + 5 = 0 \Rightarrow 6 p^{2} + 3 p + 5 = 0$
So,it is of the form ax² + bx + c = 0 where a = 6, b = 3 and c = 5.

(6) x² – 9 = 13
$\Rightarrow x^{2} - 9 - 13 = 0 \Rightarrow x^{2} - 22 = 0 \Rightarrow x^{2} + 0 x - 22 = 0$
So,it is of the form ax² + bx + c = 0 where a = 1, b = 0 and c = −22.

Page No 34:

Question 4:

Determine whether the values given against each of the quadratic equation are the roots of the equation.
(1) x² + 4x – 5 = 0 , x = 1, –1
(2) 2m² – 5m = 0, $m = 2, \frac{5}{2}$

Answer:

(1) x² + 4x – 5 = 0 , x = 1, –1
$F o r x = 1 {(1)}^{2} + 4 (1) - 5 = 0 \Rightarrow 1 + 4 - 5 = 0 \Rightarrow 5 - 5 = 0$
So, x = 1 is a solution of the given equation.
For x = –1
${(- 1)}^{2} + 4 (- 1) - 5 = 0 \Rightarrow 1 - 4 - 5 = 0 \Rightarrow - 3 - 5 = 0 \Rightarrow - 8 \neq 0$
So, x = –1 is not a solution of the given equation.
Thus, only x = 1 is root of the given equation.

(2) 2m² – 5m = 0, $m = 2, \frac{5}{2}$
$W h e n m = 2 2 {(2)}^{2} - 5 (2) = 0 \Rightarrow 8 - 10 = 0 \Rightarrow - 2 \neq 0$
So, m = 2 is not a solution of the given equation.
$When m = \frac{5}{2} 2 {(\frac{5}{2})}^{2} - 5 (\frac{5}{2}) = 0 \Rightarrow \frac{25}{2} - \frac{25}{2} = 0$
So, $m = \frac{5}{2}$ is a solution of the given quadratic equation.
Thus, only $m = \frac{5}{2}$ is a root of the given quadratic equation.

Page No 34:

Question 5:

Find k if x = 3 is a root of equation kx² – 10x + 3 = 0

Answer:

Given x = 3 is a root of equation kx² – 10x + 3 = 0
So, x = 3 must satisfy the given quadration equation.
$k {(3)}^{2} - 10 (3) + 3 = 0 \Rightarrow 9 k - 30 + 3 = 0 \Rightarrow 9 k - 27 = 0 \Rightarrow 9 k = 27 \Rightarrow k = \frac{27}{9} = 3$
Thus, k = 3.

Page No 34:

Question 6:

One of the roots of equation 5m² + 2m + k = 0 is $\frac{- 7}{5}$ . Complete the following activity to find the value of 'k'.

Answer:

$\frac{- 7}{5}$ is a root of the quadratic equation 5m² + 2m + k = 0.
∴ Put m = $\frac{- 7}{5}$ in the equation.
$5 \times {\frac{- 7}{5}}^{2} + 2 \times \frac{- 7}{5} + k = 0 \frac{49}{5} + \frac{- 14}{5} + k = 0 7 + k = 0 k = - 7$

Page No 36:

Question 1:

Solve the following quadratic equations by factorisation.
(1) x² – 15x + 54 = 0
(2) x² + x – 20 = 0
(3) 2y² + 27y + 13 = 0
(4) 5m² = 22m + 15
(5) 2x² – 2x + $\frac{1}{2}$ = 0
(6) $6 x - \frac{2}{x} = 1$
(7) $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$
to solve this quadratic equation by factorisation, complete the following activity.
(8) $3 x^{2} - 2 \sqrt{6} x + 2 = 0$
(9) $2 m (m - 24) = 50$
(10) $25 m^{2} = 9$
(11) $7 m^{2} = 21 m$
(12) $m^{2} - 11 = 0$

Answer:

(1) x² – 15x + 54 = 0
$\Rightarrow x^{2} - 9 x - 6 x + 54 = 0 \Rightarrow x (x - 9) - 6 (x - 9) = 0 \Rightarrow (x - 6) (x - 9) = 0 \Rightarrow (x - 6) = 0 or (x - 9) = 0 \Rightarrow x = 6 or x = 9$
So, 6 and 9 are the roots of the given quadratic equation.

(2) x² + x – 20 = 0
$\Rightarrow x^{2} + 5 x - 4 x - 20 = 0 \Rightarrow x (x + 5) - 4 (x + 5) = 0 \Rightarrow (x - 4) (x + 5) = 0 \Rightarrow (x - 4) = 0 or (x + 5) = 0 \Rightarrow x = 4 or x = - 5$
So, 4 and $-$ 5 are the roots of the given quadratic equation.

(3) 2y² + 27y + 13 = 0
$2 y^{2} + 26 y + y + 13 = 0 \Rightarrow 2 y (y + 13) + (y + 13) = 0 \Rightarrow (2 y + 1) (y + 13) = 0 \Rightarrow (2 y + 1) = 0 or (y + 13) = 0 \Rightarrow y = \frac{- 1}{2} or y = - 13$
So, $\frac{- 1}{2} and - 13$ are the roots of the given quadratic equation.

(4) 5m² = 22m + 15
$\Rightarrow 5 m^{2} - 22 m - 15 = 0 \Rightarrow 5 m^{2} - 25 m + 3 m - 15 = 0 \Rightarrow 5 m (m - 5) + 3 (m - 5) = 0 \Rightarrow (5 m + 3) (m - 5) = 0 \Rightarrow (5 m + 3) = 0 or (m - 5) = 0 \Rightarrow m = \frac{- 3}{5} or m = 5$
So, $\frac{- 3}{5} and 5$ are the roots of the given quadratic equation.

(5) 2x² – 2x + $\frac{1}{2}$ = 0
$\Rightarrow 4 x^{2} - 4 x + 1 = 0 \Rightarrow {(2 x)}^{2} - 2 \times 1 \times 2 x + 1 = 0 U \sin g t h e i d e n t i t y, {(a - b)}^{2} = a^{2} + b^{2} - 2 a b {(2 x)}^{2} - 2 \times 1 \times 2 x + 1 = 0 \Rightarrow {(2 x - 1)}^{2} = 0 \Rightarrow x = \frac{1}{2}, \frac{1}{2}$
So, $\frac{1}{2} and \frac{1}{2}$ are the roots of the given quadratic equation.

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

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

(8) $3 x^{2} - 2 \sqrt{6} x + 2 = 0$
$\Rightarrow 3 x^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \Rightarrow \sqrt{3} x (\sqrt{3} x - \sqrt{2}) - \sqrt{2} (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow (\sqrt{3} x - \sqrt{2}) (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}$

(9) $2 m (m - 24) = 50$
$\Rightarrow m (m - 24) = 25 \Rightarrow m^{2} - 24 m = 25 \Rightarrow m^{2} - 24 m - 25 = 0 \Rightarrow m^{2} - 25 m + m - 25 = 0 \Rightarrow m (m - 25) + 1 (m - 25) = 0 \Rightarrow (m + 1) (m - 25) = 0 \Rightarrow m = - 1, 25$

(10) $25 m^{2} = 9$
$\Rightarrow 25 m^{2} - 9 = 0 \Rightarrow {(5 m)}^{2} - 3^{2} = 0 \Rightarrow (5 m - 3) (5 m + 3) = 0 (∵ (x - a) (x + a) = x^{2} - a^{2}) \Rightarrow (5 m - 3) = 0 o r (5 m + 3) = 0 \Rightarrow m = \frac{3}{5}, \frac{- 3}{5}$

(11) $7 m^{2} = 21 m$
$\Rightarrow 7 m^{2} - 21 m = 0 \Rightarrow 7 m (m - 3) = 0 \Rightarrow 7 m = 0 or m - 3 = 0 \Rightarrow m = 0 or m = 3$

(12) $m^{2} - 11 = 0$
$\Rightarrow m^{2} - {(\sqrt{11})}^{2} = 0 \Rightarrow (m - \sqrt{11}) (m + \sqrt{11}) = 0 \Rightarrow m = \sqrt{11}, - \sqrt{11}$

Page No 39:

Question 1:

Solve the following quadratic equations by completing the square method.
(1) x² + x – 20 = 0
(2) x² + 2x – 5 = 0
(3) m² – 5m = –3
(4) 9y² – 12y + 2 = 0
(5) 2y² + 9y +10 = 0
(6) 5x² = 4x + 7

Answer:

(1) x² + x – 20 = 0
$x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 20 = 0 \Rightarrow {(x + \frac{1}{2})}^{2} = {(\frac{1}{2})}^{2} + 20 \Rightarrow {(x + \frac{1}{2})}^{2} = \frac{1}{4} + 20 \Rightarrow {(x + \frac{1}{2})}^{2} = \frac{81}{4} \Rightarrow x + \frac{1}{2} = \frac{9}{2} or x + \frac{1}{2} = - \frac{9}{2} \Rightarrow x = 4 or x = - 5$
(2) x² + 2x – 5 = 0
$\Rightarrow x^{2} + 2 x + {(\frac{2}{2})}^{2} - {(\frac{2}{2})}^{2} - 5 = 0 \Rightarrow (x^{2} + 2 x + 1) - 1 - 5 = 0 \Rightarrow {(x + 1)}^{2} - 6 = 0 \Rightarrow {(x + 1)}^{2} = 6 \Rightarrow {(x + 1)}^{2} = {(\sqrt{6})}^{2} \Rightarrow x + 1 = \sqrt{6} or x + 1 = - \sqrt{6} \Rightarrow x = \sqrt{6} - 1 or x = - \sqrt{6} - 1$

(3) m² – 5m = –3
$\Rightarrow m^{2} - 5 m + {(\frac{- 5}{2})}^{2} - {(\frac{- 5}{2})}^{2} = - 3 \Rightarrow (m^{2} - 5 m + \frac{25}{4}) - \frac{25}{4} = - 3 \Rightarrow {(m - \frac{5}{2})}^{2} = - 3 + \frac{25}{4} \Rightarrow {(m - \frac{5}{2})}^{2} = \frac{13}{4} \Rightarrow {(m - \frac{5}{2})}^{2} = {(\frac{\sqrt{13}}{2})}^{2} \Rightarrow m - \frac{5}{2} = \frac{\sqrt{13}}{2} or m - \frac{5}{2} = - \frac{\sqrt{13}}{2} \Rightarrow m = \frac{\sqrt{13}}{2} + \frac{5}{2} or m = - \frac{\sqrt{13}}{2} + \frac{5}{2} \Rightarrow m = \frac{\sqrt{13} + 5}{2} or m = \frac{5 - \sqrt{13}}{2}$

(4) 9y² – 12y + 2 = 0
$\Rightarrow y^{2} - \frac{12}{9} y + \frac{2}{9} = 0 \Rightarrow y^{2} - \frac{4}{3} y + \frac{2}{9} = 0 \Rightarrow y^{2} - \frac{4}{3} y + {(\frac{\frac{4}{3}}{2})}^{2} - {(\frac{\frac{4}{3}}{2})}^{2} + \frac{2}{9} = 0 \Rightarrow y^{2} - \frac{4}{3} y + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{2}{9} = 0$
$\Rightarrow [y^{2} - \frac{4}{3} y + (\frac{4}{9})] - (\frac{4}{9}) + \frac{2}{9} = 0 \Rightarrow {(y - \frac{2}{3})}^{2} - \frac{2}{9} = 0 \Rightarrow {(y - \frac{2}{3})}^{2} = \frac{2}{9} \Rightarrow {(y - \frac{2}{3})}^{2} = {(\frac{\sqrt{2}}{3})}^{2}$
$\Rightarrow y - \frac{2}{3} = \frac{\sqrt{2}}{3} or y - \frac{2}{3} = - \frac{\sqrt{2}}{3} \Rightarrow y = \frac{\sqrt{2}}{3} + \frac{2}{3} or y = - \frac{\sqrt{2}}{3} + \frac{2}{3} \Rightarrow y = \frac{\sqrt{2} + 2}{3} or y = \frac{- \sqrt{2} + 2}{3}$

(5) 2y² + 9y +10 = 0
$\Rightarrow y^{2} + \frac{9}{2} y + 5 = 0 \Rightarrow y^{2} + \frac{9}{2} y + {(\frac{\frac{9}{2}}{2})}^{2} - {(\frac{\frac{9}{2}}{2})}^{2} + 5 = 0 \Rightarrow [y^{2} + \frac{9}{2} y + {(\frac{9}{4})}^{2}] - {(\frac{9}{4})}^{2} + 5 = 0 \Rightarrow {(y + \frac{9}{4})}^{2} = \frac{81}{16} - 5 \Rightarrow {(y + \frac{9}{4})}^{2} = \frac{1}{16}$
$\Rightarrow {(y + \frac{9}{4})}^{2} = {(\frac{1}{4})}^{2} \Rightarrow y + \frac{9}{4} = \frac{1}{4} or y + \frac{9}{4} = - \frac{1}{4} \Rightarrow y = \frac{1}{4} - \frac{9}{4} = \frac{- 8}{4} = - 2 or y = - \frac{1}{4} - \frac{9}{4} = \frac{- 10}{4} = - \frac{5}{2} \Rightarrow y = - 2, - \frac{5}{2}$
(6) 5x² = 4x + 7
$\Rightarrow 5 x^{2} - 4 x - 7 = 0 \Rightarrow x^{2} - \frac{4}{5} x - \frac{7}{5} = 0 \Rightarrow x^{2} - \frac{4}{5} x + {(\frac{\frac{4}{5}}{2})}^{2} - {(\frac{\frac{4}{5}}{2})}^{2} - \frac{7}{5} = 0 \Rightarrow [x^{2} - \frac{4}{5} x + {(\frac{2}{5})}^{2}] - {(\frac{2}{5})}^{2} - \frac{7}{5} = 0 \Rightarrow {(x - \frac{2}{5})}^{2} = \frac{7}{5} + \frac{4}{25} = \frac{35 + 4}{25} = \frac{39}{25} \Rightarrow {(x - \frac{2}{5})}^{2} = \frac{39}{25}$
$\Rightarrow {(x - \frac{2}{5})}^{2} = {(\frac{\sqrt{39}}{5})}^{2} \Rightarrow x - \frac{2}{5} = \frac{\sqrt{39}}{5} or x - \frac{2}{5} = - \frac{\sqrt{39}}{5} \Rightarrow x = \frac{2 + \sqrt{39}}{5} or x = \frac{2 - \sqrt{39}}{5}$

Page No 43:

Question 1:

Compare the given quadratic equations to the general form and write values of a,b, c.
(1) x² – 7x + 5 = 0
(2) 2m² = 5m – 5
(3) y² = 7y

Answer:

(1) x² – 7x + 5 = 0
General form of the quadratic equation is $a x^{2} + b x + c = 0$ .
Comparing x² – 7x + 5 = 0 with the general form we have a = 1, b = $-$ 7 and c = 5.

(2) 2m² = 5m – 5
$\Rightarrow 2 m^{2} - 5 m + 5 = 0$
General form of the quadratic equation is $a x^{2} + b x + c = 0$ .
Comparing 2m² = 5m – 5 with the general form we have a = 2, b = $-$ 5 and c = 5.

(3) y² = 7y
$\Rightarrow y^{2} - 7 y + 0 = 0$
General form of the quadratic equation is $a x^{2} + b x + c = 0$ .
Comparing $y^{2} - 7 y + 0 = 0$ with the general form we have a = 1, b = $-$ 7 and c = 0.

Page No 43:

Question 2:

Solve using formula.
(1) x² + 6x + 5 = 0
(2) x² – 3x – 2 = 0
(3) 3m² + 2m – 7 = 0
(4) 5m² – 4m – 2 = 0
(5) $y^{2} + \frac{1}{3} y = 2$
(6) 5x² + 13x + 8 = 0

Answer:

(1) x² + 6x + 5 = 0
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 1, b = 6 and c = 5
Now $b^{2} - 4 a c = 6^{2} - 4 \times 1 \times 5 = 36 - 20 = 16$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- 6 \pm \sqrt{16}}{2 \times 1} = \frac{- 6 \pm 4}{2} \Rightarrow x = \frac{- 6 + 4}{2} or x = \frac{- 6 - 4}{2} \Rightarrow x = - 1 or x = - 5$

(2) x² – 3x – 2 = 0
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 1, b = $-$ 3 and c = $-$ 2
Now $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times (- 2) = 9 + 8 = 17$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{3 \pm \sqrt{17}}{2 \times 1} = \frac{3 \pm \sqrt{17}}{2} \Rightarrow x = \frac{3 + \sqrt{17}}{2} or x = \frac{3 - \sqrt{17}}{2}$

(3) 3m² + 2m – 7 = 0
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 3, b = 2 and c = $-$ 7
Now $b^{2} - 4 a c = {(2)}^{2} - 4 \times 3 \times (- 7) = 4 + 84 = 88$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- 2 \pm \sqrt{88}}{2 \times 3} = \frac{- 2 \pm \sqrt{88}}{6} \Rightarrow x = \frac{- 2 + \sqrt{88}}{6} or x = \frac{- 2 - \sqrt{88}}{6} \Rightarrow x = \frac{- 1 + \sqrt{22}}{3} or \frac{- 1 - \sqrt{22}}{3}$

(4) 5m² – 4m – 2 = 0
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 5, b = – 4and c = $-$ 2
Now $b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 5 \times (- 2) = 16 + 40 = 56$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- 4 \pm \sqrt{56}}{2 \times 5} = \frac{- 4 \pm \sqrt{56}}{10} \Rightarrow x = \frac{- 4 + \sqrt{56}}{10} or x = \frac{- 4 - \sqrt{56}}{10} \Rightarrow x = \frac{- 2 + \sqrt{14}}{5} or \frac{- 2 - \sqrt{14}}{5}$

(5) $y^{2} + \frac{1}{3} y = 2$
Multiplying the equation by 3
$3 y^{2} + y = 6 \Rightarrow 3 y^{2} + y - 6 = 0$
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 3, b = 1 and c = $-$ 6
Now $b^{2} - 4 a c = {(1)}^{2} - 4 \times 3 \times (- 6) = 1 + 72 = 73$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- 1 \pm \sqrt{73}}{2 \times 3} = \frac{- 1 \pm \sqrt{73}}{6} \Rightarrow x = \frac{- 1 + \sqrt{73}}{6} or x = \frac{- 1 - \sqrt{73}}{6}$

(6) 5x² + 13x + 8 = 0
On comparing with the equation $a x^{2} + b x + c = 0$ ,
a = 5, b = 13 and c = 8
Now $b^{2} - 4 a c = {(13)}^{2} - 4 \times 5 \times 8 = 169 - 160 = 9$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{- 13 \pm \sqrt{9}}{2 \times 5} = \frac{- 13 \pm 3}{10} \Rightarrow x = \frac{- 13 + 3}{10} or x = \frac{- 13 - 3}{10} \Rightarrow x = \frac{- 10}{10} or x = \frac{- 16}{10} \Rightarrow x = - 1 or x = \frac{- 8}{5}$

Page No 44:

Question 3:

With the help of the flow chart given below solve the equation $x^{2} + 2 \sqrt{3} x + 3 = 0$ using the formula.

Answer:

$Compare equations x^{2} + 2 \sqrt{3} x + 3 = 0 and a x^{2} + b x + c = 0 find the values of a, b, c . \to Find value of b^{2} - 4 a c \to Write formula to solve quadratic equation . \to Substitute values of a, b, c and find roots$

comparing $x^{2} + 2 \sqrt{3} x + 3 = 0$ with $a x^{2} + b x + c = 0$ we get
$a = 1, b = 2 \sqrt{3} and c = 3$
$b^{2} - 4 a c = {(2 \sqrt{3})}^{2} - 4 \times 1 \times 3 = 12 - 12 = 0$
Formula to solve a quadratic equation will be
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- 2 \sqrt{3} \pm \sqrt{0}}{2 \times 1} = \frac{- 2 \sqrt{3}}{2} = - \sqrt{3}$
Thus, $x = - \sqrt{3}, - \sqrt{3}$ .

Page No 49:

Question 1:

Fill in the gaps and complete.

(3)If α, β are roots of quadratic equation,

Answer:

If $α and β$ are the roots of the quadratic equation then the quadratic equation can be written as
$x^{2} - (α + β) x + α β = 0$
So, the quadratic equation with sum of roots as $α + β = - 7$ and product of roots as $α β = 5$ will be
$x^{2} + 7 x + 5 = 0$

Sum of roots = $\frac{- b}{a} = \frac{- (- 4)}{2} = 2$
Product of roots = $\frac{c}{a} = \frac{- 3}{2}$

Page No 49:

Question 2:

Find the value of discriminant.
(1) x² + 7x – 1 = 0
(2) 2y² – 5y + 10 = 0
(3) $\sqrt{2} x^{2} + 4 x + 2 \sqrt{2} = 0$

Answer:

(1) x² + 7x – 1 = 0
Comparing the given equation with $a x^{2} + b x + c = 0$ ,
$a = 1, b = 7 and c = - 1$
So, the discriminant $b^{2} - 4 a c = {(7)}^{2} - 4 \times 1 \times (- 1) = 49 + 4 = 53$

(2) 2y² – 5y + 10 = 0
Comparing the given equation with $a x^{2} + b x + c = 0$ ,
$a = 2, b = - 5 and c = 10$
So, the discriminant $b^{2} - 4 a c = {(- 5)}^{2} - 4 \times 2 \times 10 = 25 - 80 = - 55$

(3) $\sqrt{2} x^{2} + 4 x + 2 \sqrt{2} = 0$
Comparing the given equation with $a x^{2} + b x + c = 0$ ,
$a = \sqrt{2}, b = 4 and c = 2 \sqrt{2}$
So, the discriminant $b^{2} - 4 a c = {(4)}^{2} - 4 \times \sqrt{2} \times 2 \sqrt{2} = 16 - 16 = 0$

Page No 49:

Question 3:

Determine the nature of roots of the following quadratic equations.
(1) x² – 4x + 4 = 0
(2) 2y²– 7y +2 = 0
(3) m² + 2m + 9 = 0

Answer:

(1) x² – 4x + 4 = 0
Comparing the given equation with the quadratic equation $a x^{2} + b x + c = 0$ ,
$a = 1, b = - 4, c = 4$
Discriminant, $△ = b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 1 \times 4 = 16 - 16 = 0$
Since the discriminant = 0 so, the roots of the given quadratic equation are real and equal.

(2) 2y²– 7y +2 = 0
Comparing the given equation with the quadratic equation $a x^{2} + b x + c = 0$ ,
$a = 2, b = - 7, c = 2$
Discriminant, $△ = b^{2} - 4 a c = {(- 7)}^{2} - 4 \times 2 \times 2 = 49 - 16 = 33$
Since the discriminant > 0 so, the roots of the given quadratic equation are real and unequal.

(3) m² + 2m + 9 = 0
Comparing the given equation with the quadratic equation $a x^{2} + b x + c = 0$ ,
$a = 1, b = 2, c = 9$
Discriminant, $△ = b^{2} - 4 a c = {(2)}^{2} - 4 \times 1 \times 9 = 4 - 36 = - 32$
Since the discriminant < 0 so, the roots of the given quadratic equation are not real.

Page No 50:

Question 4:

Form the quadratic equation from the roots given below.
(1) 0 and 4
(2) 3 and –10
(3) $\frac{1}{2}, - \frac{1}{2}$
(4) $2 - \sqrt{5}, 2 + \sqrt{5}$

Answer:

(1) 0 and 4
Sum of roots = 0 + 4 = 4
Product of roots = 0 $\times$ 4 = 0
The general form of the quadratic equation is $x^{2} - (Sum of roots) x + Product of roots = 0$
So, the quadratic equation obtained is $x^{2} - 4 x + 0 = 0$
$\Rightarrow x^{2} - 4 x = 0 \Rightarrow x (x - 4) = 0$

(2) 3 and –10
Sum of roots = 3 + (–10) = −7
Product of roots = 3 $\times$ –10 = –30
The general form of the quadratic equation is $x^{2} - (Sum of roots) x + Product of roots = 0$
So, the quadratic equation obtained is $x^{2} - (- 7) x + (- 30) = 0$
$x^{2} + 7 x - 30 = 0$

(3) $\frac{1}{2}, - \frac{1}{2}$
Sum of roots = $\frac{1}{2} + (- \frac{1}{2}) = 0$
Product of roots = $\frac{1}{2} \times (\frac{- 1}{2}) = \frac{- 1}{4}$
The general form of the quadratic equation is $x^{2} - (Sum of roots) x + Product of roots = 0$
So, the quadratic equation obtained is $x^{2} - (0) x + (\frac{- 1}{4}) = 0$
$\Rightarrow x^{2} - \frac{1}{4} = 0$

(4) $2 - \sqrt{5}, 2 + \sqrt{5}$
Sum of roots = $2 - \sqrt{5} + 2 + \sqrt{5} = 4$
Product of roots = $(2 - \sqrt{5}) (2 + \sqrt{5}) = 4 - 5 = - 1$
The general form of the quadratic equation is $x^{2} - (Sum of roots) x + Product of roots = 0$
So, the quadratic equation obtained is $x^{2} - (4) x + (- 1) = 0$
$\Rightarrow x^{2} - 4 x - 1 = 0$

Page No 50:

Question 5:

Sum of the roots of a quadratic equation is double their product. Find k if equation x² – 4kx + k +3 = 0

Answer:

Let the roots of the quadratic equation be $α and β$ .
Sum of roots = $α + β$
Product of roots = $α β$
Given that sum of roots = double the product
$\Rightarrow α + β = 2 α β$ .....(I)
Given: x² – 4kx + k +3 = 0 .....(II)
General form of the quadratic equation is $x^{2} - (Sum of roots) x + Product of roots = 0$
$\Rightarrow x^{2} - (α + β) x + α β = 0 \Rightarrow x^{2} - (2 α β) x + α β = 0 . . . . . (III)$
On comparing (II) and (III) we have
$4 k = 2 α β and k + 3 = α β$
Solving these equations we have
$4 k = 2 (k + 3) \Rightarrow 2 k = k + 3 \Rightarrow 2 k - k = 3 \Rightarrow k = 3$

Page No 50:

Question 6:

α, β are roots of y² – 2y –7 = 0 find,
(1) α² + β²
(2) α³ + β³

Answer:

α, β are roots of y² – 2y –7 = 0
$a = 1, b = - 2, c = - 7$ $a = 1, b = - 2, c = - 7$
(1)
$α + β = \frac{- b}{a} = \frac{- (- 2)}{1} = 2 α β = \frac{c}{a} = \frac{- 7}{1} = - 7$
$α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = {(2)}^{2} - 2 (- 7) = 4 + 14 = 18$

(2)
$α^{3} + β^{3} = {(α + β)}^{3} - 3 α β (α + β) = {(2)}^{3} - 3 (- 7) (2) = 8 + 42 = 50$

Page No 50:

Question 7:

The roots of each of the following quadratic equations are real and equal, find k.
(1) 3y²+ ky +12 = 0
(2) kx (x – 2) + 6 = 0

Answer:

(1) 3y²+ ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
$b^{2} - 4 a c = 0 \Rightarrow k^{2} - 4 \times 3 \times 12 = 0 \Rightarrow k^{2} - 144 = 0 \Rightarrow k^{2} = 144 \Rightarrow k = \pm 12$

(2) kx (x – 2) + 6 = 0
$\Rightarrow k x^{2} - 2 k x + 6 = 0$
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
$b^{2} - 4 a c = 0 \Rightarrow {(- 2 k)}^{2} - 4 \times k \times 6 = 0 \Rightarrow 4 k^{2} - 24 k = 0 \Rightarrow 4 k (k - 6) = 0 \Rightarrow 4 k = 0 or k - 6 = 0 \Rightarrow k = 0 or k = 6$
But k cannot be equal to 0 since then there will not be any quadratic equation.
So, k = 6

Page No 52:

Question 1:

Product of Pragati’s age 2 years ago and 3 years hence is 84. Find her present age.

Answer:

Let Pragati's present age be x years.
Her age 2 years ago = $x - 2$
Her age 3 years hence = $x + 3$
Product of Pragati’s age 2 years ago and 3 years hence is 84.
$(x - 2) (x + 3) = 84 \Rightarrow x^{2} + x - 6 = 84 \Rightarrow x^{2} + x - 90 = 0 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} - 4 \times 1 \times (- 90)}}{2} = \frac{- 1 \pm \sqrt{361}}{2} x = \frac{- 1 - 19}{2} or \frac{- 1 + 19}{2} \Rightarrow x = - 10 or 9$
But age cannot be negative so, x = 9.
Thus, Pragati's present age is 9 years.

Page No 52:

Question 2:

The sum of squares of two consecutive natural numbers is 244; find the numbers.

Answer:

Disclaimer: There is an error in the given question. Here it should be two consecutive even natural numbers.

Let the two consecutive natural numbers be $x and x + 2$ .
Sum of squares of these two consecutive natural numbers is 244
$x^{2} + {(x + 2)}^{2} = 244 \Rightarrow x^{2} + x^{2} + 4 + 4 x = 244 \Rightarrow 2 x^{2} + 4 x - 240 = 0 \Rightarrow x^{2} + 2 x - 120 = 0 \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times (- 120)}}{2} \Rightarrow x = \frac{- 2 \pm \sqrt{4 + 480}}{2}$
$\Rightarrow x = \frac{- 2 \pm \sqrt{484}}{2} \Rightarrow x = \frac{- 2 \pm 22}{2} \Rightarrow x = \frac{- 2 + 22}{2}, \frac{- 2 - 22}{2} \Rightarrow x = \frac{20}{2}, \frac{- 24}{2} \Rightarrow x = 10, - 12$
But the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.

Page No 52:

Question 3:

In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.

Answer:

Let the number of trees in each column be x.
Number of trees in a row = x + 5
Total trees = Number of rows $\times$ Number of columns = $x (x + 5)$ = 150
$\Rightarrow x (x + 5) = 150 \Rightarrow x^{2} + 5 x - 150 = 0 \Rightarrow x^{2} + 15 x - 10 x - 150 = 0 \Rightarrow x (x + 15) - 10 (x + 15) = 0 \Rightarrow (x + 15) (x - 10) = 0 \Rightarrow x = 10, - 15$
But number of columns cannot be negative so, number of columns = 10.
Number of rows = 10 + 5 = 15

Page No 52:

Question 4:

Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is $\frac{1}{6}$ . Find their present ages.

Answer:

Let the present age of Kishor be x years.
Vivek is older than Kishor by 5 years.
So, Vivek's age will be x + 5.
The sum of the reciprocals of their ages is $\frac{1}{6}$ .
$\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6} \Rightarrow \frac{x + 5 + x}{x (x + 5)} = \frac{1}{6} \Rightarrow \frac{5 + 2 x}{x^{2} + 5 x} = \frac{1}{6} \Rightarrow 30 + 12 x = x^{2} + 5 x \Rightarrow x^{2} - 7 x - 30 = 0 \Rightarrow x^{2} - 10 x + 3 x - 30 = 0 \Rightarrow x (x - 10) + 3 (x - 10) = 0 \Rightarrow (x + 3) (x - 10) = 0 \Rightarrow x = - 3, 10$
But age cannot be negative so, age of Kishore is 10 years and that of Vivek is 10 + 5 = 15 years.

Page No 52:

Question 5:

Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.

Answer:

Let the score in the first test be x.
Suyash's score in second test is 10 more than that in the first.
So, his score in the second test = x + 10
5 times the score of the second test is the same as square of the score in the first test.
$5 (x + 10) = x^{2} \Rightarrow 5 x + 50 = x^{2} \Rightarrow x^{2} - 5 x - 50 = 0 \Rightarrow x^{2} - 10 x + 5 x - 50 = 0 \Rightarrow x (x - 10) + 5 (x - 10) = 0 \Rightarrow (x + 5) (x - 10) = 0 \Rightarrow x = 10, - 5$
The score is not negative so, Suyash scored 10 marks in first test.

Page No 52:

Question 6:

Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is Rs 600, find production cost of one pot and number of pots he makes per day.

Answer:

Let the number of pots Mr. Kasam makes in one day be x.
Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day = 10x + 40
Production cost of all pots per day is Rs 600
(10x + 40)x = 600
$10 x^{2} + 40 x = 600 \Rightarrow x^{2} + 4 x - 60 = 0 \Rightarrow x^{2} + 10 x - 6 x - 60 = 0 \Rightarrow x (x + 10) - 6 (x + 10) = 0 \Rightarrow (x - 6) (x + 10) = 0 \Rightarrow x = 6, - 10$
But the number of pots cannot be negative so,
$x \neq - 10 \Rightarrow x = 6$
Production cost of 1 pot = $10 \times 6 + 40 = 60 + 40 = Rs 100$

Page No 52:

Question 7:

Pratik takes 8 hours to travel 36 km down stream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.

Answer:

Let the speed of water current = y km/h
Speed of boat in still water = 12 km/h
Speed upstream = $12 - y$
Speed downstream = 12 + y
Distance travelled = 36 km downstream and 36 km upstream
$Speed = \frac{Distance}{Time} \Rightarrow Time = \frac{Distance}{Speed}$
$\frac{36}{12 - y} + \frac{36}{12 + y} = 8 \Rightarrow 36 [\frac{1}{12 - y} + \frac{1}{12 + y}] = 8 \Rightarrow \frac{12 + y + 12 - y}{144 - y^{2}} = \frac{1}{4} \Rightarrow \frac{24}{144 - y^{2}} = \frac{1}{4} \Rightarrow 144 - y^{2} = 24 \times 4 \Rightarrow y^{2} = 36 \Rightarrow y^{2} - 36 = 0 \Rightarrow y = \pm 6$
Speed cannot be negative so, the speed of water current = 6 km/h.

Page No 52:

Question 8:

Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.

Answer:

Let the number of days taken by Nishu be x.
So, number of days taken by Pintu will be x + 6
Nishu's 1 day work = $\frac{1}{x}$
Pintu's one day work = $\frac{1}{x + 6}$
Work done together in 4 days = $\frac{1}{4}$
$\frac{1}{x} + \frac{1}{x + 6} = \frac{1}{4} \Rightarrow \frac{x + 6 + x}{x (x + 6)} = \frac{1}{4} \Rightarrow \frac{6 + 2 x}{x^{2} + 6 x} = \frac{1}{4} \Rightarrow 24 + 8 x = x^{2} + 6 x \Rightarrow x^{2} - 2 x - 24 = 0 \Rightarrow x^{2} - 6 x + 4 x - 24 = 0 \Rightarrow (x - 6) (x + 4) = 0 \Rightarrow x = 6, - 4$
Number of days cannot be negative so, x = 6.
Thus, number of days taken by Nishu = 6 and that by Pintu = 6 + 6 = 12.

Page No 52:

Question 9:

If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and diviser.

Answer:

Let the divisor be x.
Dividend = 460
Quotient = 5x + 6
Remainder = 1
We know
$Divident = Divisor \times Quotient + Remainder$
$460 = x (5 x + 6) + 1 \Rightarrow 460 = 5 x^{2} + 6 x + 1 \Rightarrow 5 x^{2} + 6 x - 459 = 0 \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- 6 \pm \sqrt{{(6)}^{2} - 4 \times 5 \times (- 459)}}{2 \times 5} \Rightarrow x = \frac{- 6 \pm \sqrt{36 + 9180}}{10} \Rightarrow x = \frac{- 6 \pm \sqrt{9216}}{10}$
$\Rightarrow x = \frac{- 6 \pm 96}{10} \Rightarrow x = \frac{- 6 + 96}{10}, \frac{- 6 - 96}{10} \Rightarrow x = 9, - 10.2$
Thus, divisor = 9, quotient = 5x + 6 = $5 \times 9 + 6 = 45 + 6 = 51$

Page No 52:

Question 10:

In the adjoining fig. $□$ ABCD is a trapezium AB || CD and its area is 33 cm² . From the information given in the figure find the lengths of all sides of the $□$ ABCD. Fill in the empty boxes to get the solution.

Answer:

$□$ ABCD is a trapezium.
AB || CD
A( $□$ ABCD) = $\frac{1}{2} (AB + CD) \times AM$
$33 = \frac{1}{2} (x + 2 x + 1) \times (x - 4) ∴ 33 \times 2 = (3 x + 1) \times (x - 4) ∴ 3 x^{2} + 11 x - 70 = 0 ∴ 3 x (x - 7) + 10 (x - 7) = 0 ∴ (3 x + 10) (x - 7) = 0 ∴ (3 x + 10) = 0 or x - 7 = 0 ∴ x = \frac{- 10}{3} or x = 7$
But length is never negative.
$∴ x \neq \frac{- 10}{3} ∴ x = 7$
AB = 7, CD = 15, BC = 5

Page No 53:

Question 1:

Choose the correct answers for the following questions.
(1) Which one is the quadratic equation ?

(A)

\frac{5}{x} - 3 = x^{2}

(B) x(x + 5) = 2

(D)

\frac{1}{x^{2}} (x + 2) = x

(2) Out of the following equations which one is not a quadratic equation ?

(A) x² + 4x = 11 + x²

(B) x² = 4x

(D) 2x – x² = x² + 5

(3) The roots of x² + kx + k = 0 are real and equal, find k.

(A) 0

(B) 4

(D) 2

(4) For

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

find the value of the discriminant.

(A) –5

(B) 17

(C)

\sqrt{2}

(D)

2 \sqrt{2} - 5

(5) Which of the following quadratic equations has roots 3,5 ?

(A) x² – 15x + 8 = 0	(B) x² – 8x + 15 = 0
(C) x² + 3x + 5 = 0	(D) x² + 8x – 15 = 0

(6) Out of the following equations, find the equation having the sum of its roots –5.

(A) 3x² – 15x + 3 = 0	(B) x² – 5x + 3 = 0
(C) x² + 3x – 5 = 0	(D) 3x² + 15x + 3 = 0

(7)

\sqrt{5} m^{2} - \sqrt{5} m + \sqrt{5} = 0

which of the following statement is true for this given equation ?

(A) Real and uneual roots	(B) Real and equal roots
(C) Roots are not real	(D) Three roots.

(8) One of the roots of equation x² + mx – 5 = 0 is 2; find m.

(A) –2

(B)

- \frac{1}{2}

(C)

\frac{1}{2}

(D) 2

Answer:

(1) The general form of a quadratic equation is $a x^{2} + b x + c = 0$ .
For (A), $\frac{5}{x} - 3 = x^{2}$
$5 - 3 x = x^{3}$
Thus, (A) is not quadratic equation.
For (B) x(x + 5) = 2
$x^{2} + 5 x - 2 = 0$
Thus, (B) is a qudratic equation.
(C) is also not a quadratic equation.
(D) $\frac{1}{x^{2}} (x + 2) = x$
$x + 2 = x^{3}$ which is also not quadratic.
Hence, the correct answer is option (B).

(2) (A) x² + 4x = 11 + x²
$\Rightarrow 4 x = 11$
Thus, (A) is not a quadratic equation.
(B) and (C) can be written as $x^{2} - 4 x + 0 = 0$ and $5 x^{2} - 90 + 0 = 0$ respectively.
So, B and C are quadratic equations.
(D) 2x – x² = x² + 5 can be written as $2 x^{2} - 2 x + 5 = 0$ .
So, it also forms a quadratic equation.
Hence, the correct answer is option (A).

(3) Given quadratic equation is x² + kx + k = 0
For real and equal roots, D = 0
$b^{2} - 4 a c = 0 \Rightarrow k^{2} - 4 \times 1 \times k = 0 \Rightarrow k^{2} - 4 k = 0 \Rightarrow k (k - 4) = 0 \Rightarrow k = 0, k = 4$
Hence, the correct answer is option C.

(4) $\sqrt{2} x^{2} - 5 x + \sqrt{2} = 0$
$a = \sqrt{2}, b = - 5, c = \sqrt{2} D = b^{2} - 4 a c = {(- 5)}^{2} - 4 \times \sqrt{2} \times \sqrt{2} = 25 - 8 = 17$
Hence, the correct answer is option B.

(5) The roots of the quadratic equation x² – 8x + 15 = 0 are
$x^{2} - 5 x - 3 x + 15 = 0 \Rightarrow x (x - 5) - 3 (x - 5) = 0 \Rightarrow (x - 3) (x - 5) = 0 \Rightarrow x = 3, 5$
Hence, the correct answer is option B.

(6) For the equation 3x² + 15x + 3 = 0
$Sum of roots = \frac{- b}{a} = \frac{- 15}{3} = - 5$
Hence, the correct answer is option D.

(7) For the given quadratic equation $\sqrt{5} m^{2} - \sqrt{5} m + \sqrt{5} = 0$
$D = b^{2} - 4 a c = {(- \sqrt{5})}^{2} - 4 \times \sqrt{5} \times \sqrt{5} = 5 - 20 = - 15$
Since D < 0 so, the roots are not real.
Hence, the correct answer is option C.

(8) For the equation, x² + mx – 5 = 0
one of the roots is 2. So, 2 should satisfy the given equation.
${(2)}^{2} + m \times (2) - 5 = 0 \Rightarrow 4 + 2 m - 5 = 0 \Rightarrow 2 m - 1 = 0 \Rightarrow m = \frac{1}{2}$
Hence, the correct answer is option C.

Page No 54:

Question 2:

Which of the following equations is quadratic ?
(1) $x^{2} + 2 x + 11 = 0$
(2) $x^{2} - 2 x + 5 = x^{2}$
(3) ${(x + 2)}^{2} = 2 x^{2}$

Answer:

General form of a quadratic equation is
$a x^{2} + b x + c = 0$
(1) $x^{2} + 2 x + 11 = 0$ is in the form of the general quadratic equation.
(2) $x^{2} - 2 x + 5 = x^{2}$
$\Rightarrow - 2 x + 5 = 0$
It is not in the form of quadratic equation.
(3) ${(x + 2)}^{2} = 2 x^{2}$
$\Rightarrow x^{2} + 4 + 4 x = 2 x^{2} \Rightarrow x^{2} - 4 x - 4 = 0$
It is also in the form of a quadratic equation.
Hence, (1) and (3) are quadratic equations.

Page No 54:

Question 3:

Find the value of discriminant for each of the following equations.
(1) $2 y^{2} - y + 2 = 0$
(2) $5 m^{2} - m = 0$
(3) $\sqrt{5} x^{2} - x - \sqrt{5} = 0$

Answer:

(1) $2 y^{2} - y + 2 = 0$
For the quadratic equation, $a x^{2} + b x + c = 0$
$D = b^{2} - 4 a c$
Here,
$a = 2, b = - 1, c = 2 D = {(- 1)}^{2} - 4 \times 2 \times 2 = 1 - 16 = - 15$

(2) $5 m^{2} - m = 0$
For the quadratic equation, $a x^{2} + b x + c = 0$
$D = b^{2} - 4 a c$
Here,
$m = 5, b = - 1, c = 0 D = {(- 1)}^{2} - 4 \times 5 \times 0 = 1$

(3) $\sqrt{5} x^{2} - x - \sqrt{5} = 0$
For the quadratic equation, $a x^{2} + b x + c = 0$
$D = b^{2} - 4 a c$
Here,
$a = \sqrt{5}, b = - 1, c = - \sqrt{5} D = {(- 1)}^{2} - 4 \times (\sqrt{5}) \times (- \sqrt{5}) = 1 + 20 = 21$

Page No 54:

Question 4:

One of the roots of quadratic equation $2 x^{2} + k x - 2 = 0$ is –2. find k.

Answer:

$2 x^{2} + k x - 2 = 0$
One of the roots is –2 so, it will satisfy the given equation.
$2 {(- 2)}^{2} + k (- 2) - 2 = 0 \Rightarrow 8 - 2 k - 2 = 0 \Rightarrow 6 - 2 k = 0 \Rightarrow 6 = 2 k \Rightarrow k = 3$

Page No 54:

Question 5:

Two roots of quadratic equations are given ; frame the equation.
(1) 10 and –10
(2) $1 - 3 \sqrt{5} and 1 + 3 \sqrt{5}$
(3) 0 and 7

Answer:

(1) 10 and –10
Sum of roots = 10 + (–10) = 0
Product of roots = $10 \times (- 10) = - 100$
The general form of the quadratic equation is
$x^{2} - (Sum of roots) x + product of roots = 0$
So, the quadratic equation will be
$x^{2} - 0 x - 100 = 0 \Rightarrow x^{2} - 100 = 0$

(2) $1 - 3 \sqrt{5} and 1 + 3 \sqrt{5}$
Sum of roots = $1 - 3 \sqrt{5} + 1 + 3 \sqrt{5} = 2$
Product of roots = $(1 - 3 \sqrt{5}) (1 + 3 \sqrt{5}) = 1 - 45 = - 44$
The general form of the quadratic equation is
$x^{2} - (Sum of roots) x + product of roots = 0$
So, the quadratic equation will be
$x^{2} - 2 x - 44 = 0$

(3) 0 and 7
Sum of roots = 0 + 7 = 7
Product of roots = $0 \times 7 = 0$
The general form of the quadratic equation is
$x^{2} - (Sum of roots) x + product of roots = 0$
So, the quadratic equation will be
$x^{2} - 7 x + 0 = 0 \Rightarrow x^{2} - 7 x = 0$

Page No 54:

Question 6:

Determine the nature of roots for each of the quadratic equation.
(1) $3 x^{2} - 5 x + 7 = 0$
(2) $\sqrt{3} x^{2} + \sqrt{2} x - 2 \sqrt{3} = 0$
(3) $m^{2} - 4 x - 3 = 0$

Answer:

(1) $3 x^{2} - 5 x + 7 = 0$
In order to find the nature of the roots we need to find the discriminant.
$D = b^{2} - 4 a c a = 3, b = - 5, c = 7 D = {(- 5)}^{2} - 4 \times 3 \times 7 = 25 - 84 = - 59$
Since, the D < 0 so, there is no real root.

(2) $\sqrt{3} x^{2} + \sqrt{2} x - 2 \sqrt{3} = 0$
In order to find the nature of the roots we need to find the discriminant.
$D = b^{2} - 4 a c a = \sqrt{3}, b = \sqrt{2}, c = - 2 \sqrt{3} D = {(\sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 2 + 24 = 26 > 0$
Since, the D > 0 so, real and unequal roots.

(3) $m^{2} - 4 x - 3 = 0$
In order to find the nature of the roots we need to find the discriminant.
$D = b^{2} - 4 a c a = 1, b = - 4, c = - 3 D = {(- 4)}^{2} - 4 \times (1) \times (- 3) = 16 + 12 = 28$
Since, the D > 0 so, real and unequal roots.

Page No 54:

Question 7:

Solve the following quadratic equations.
(1) $\frac{1}{x + 5} = \frac{1}{x^{2}}$
(2) $x^{2} - \frac{3 x}{10} - \frac{1}{10} = 0$
(3) ${(2 x + 3)}^{2} = 25$
(4) $m^{2} + 5 m + 5 = 0$
(5) $5 m^{2} + 2 m + 1 = 0$
(6) $x^{2} - 4 x - 3 = 0$

Answer:

(1) $\frac{1}{x + 5} = \frac{1}{x^{2}}$
$\Rightarrow x^{2} = x + 5 \Rightarrow x^{2} - x - 5 = 0 \Rightarrow x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 1 \times (- 5)}}{2 \times 1} \Rightarrow x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$
(2) $x^{2} - \frac{3 x}{10} - \frac{1}{10} = 0$
$\Rightarrow 10 x^{2} - 3 x - 1 = 0 \Rightarrow x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 \times 10 \times (- 1)}}{2 \times 10} \Rightarrow x = \frac{3 \pm \sqrt{9 + 40}}{20} = \frac{3 \pm \sqrt{49}}{20} \Rightarrow x = \frac{3 \pm 7}{20} \Rightarrow x = \frac{3 + 7}{20}, \frac{3 - 7}{20} \Rightarrow x = \frac{10}{20}, \frac{- 4}{20} \Rightarrow x = \frac{1}{2}, \frac{- 1}{5}$

(3) ${(2 x + 3)}^{2} = 25$
$\Rightarrow 4 x^{2} + 9 + 12 x = 25 \Rightarrow 4 x^{2} + 12 x - 16 = 0 \Rightarrow x^{2} + 3 x - 4 = 0 \Rightarrow x^{2} + 4 x - x - 4 = 0 \Rightarrow x (x + 4) - 1 (x + 4) = 0 \Rightarrow (x - 1) (x + 4) = 0 \Rightarrow x = 1, - 4$

(4) $m^{2} + 5 m + 5 = 0$
$\Rightarrow m = \frac{- 5 \pm \sqrt{{(5)}^{2} - 4 \times 1 \times 5}}{2} \Rightarrow m = \frac{- 5 \pm \sqrt{25 - 20}}{2} \Rightarrow m = \frac{- 5 \pm \sqrt{5}}{2}$

(5) $5 m^{2} + 2 m + 1 = 0$
$\Rightarrow m = \frac{- 2 \pm \sqrt{{(2)}^{2} - 4 \times 1 \times 5}}{2} \Rightarrow m = \frac{- 2 \pm \sqrt{4 - 20}}{2} \Rightarrow m = \frac{- 2 \pm \sqrt{- 16}}{2}$
So, the roots are not real as the discriminant is negative.

(6) $x^{2} - 4 x - 3 = 0$
$\Rightarrow x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \times 1 \times (- 3)}}{2 \times 1} \Rightarrow x = \frac{4 \pm \sqrt{16 + 12}}{2} \Rightarrow x = \frac{4 \pm \sqrt{28}}{2} \Rightarrow x = \frac{4 \pm 2 \sqrt{7}}{2} \Rightarrow x = 2 \pm \sqrt{7}$

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Question 8:

Find m if (m – 12) x² + 2(m –12) x + 2 = 0 has real and equal roots.

Answer:

For real and equal roots, the Discriminant should be 0.
(m – 12) x² + 2(m –12) x + 2 = 0
$∆ = b^{2} - 4 a c = 0 a = m - 12, b = 2 (m - 12), c = 2$
${[2 (m - 12)]}^{2} - 4 (m - 12) \times 2 = 0 \Rightarrow 4 m^{2} + 576 - 96 m - 8 m + 96 = 0 \Rightarrow 4 m^{2} - 104 m + 672 = 0 \Rightarrow m^{2} - 26 m + 168 = 0 \Rightarrow m^{2} - 14 m - 12 m + 168 = 0 \Rightarrow m (m - 14) - 12 (m - 14) = 0 \Rightarrow (m - 14) (m - 12) = 0 m = 12, 14$
But if m = 12, the quadratic equation will be formed hence, the vaalue of m is 14.

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Question 9:

The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.

Answer:

Let the roots be $α and β$ .
Sum of roots = 5
$α + β = 5$
Also,
$α^{3} + β^{3} = 35 \Rightarrow α^{3} + β^{3} = {(α + β)}^{3} - 3 α β (α + β) \Rightarrow 35 = 5^{3} - 3 α β (5) \Rightarrow 15 α β = 125 - 35 \Rightarrow α β = 6$
The general quadratic equation is
$x^{2} - (α + β) x + α β = 0$
So, the required quadratic equation will be
$x^{2} - (5) x + 6 = 0 \Rightarrow x^{2} - 5 x + 6 = 0$

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Question 10:

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation $2 x^{2} + 2 (p + q) x + p^{2} + q^{2} = 0$

Answer:

$2 x^{2} + 2 (p + q) x + p^{2} + q^{2} = 0$
Let the roots of the given quadratic equation be $α$ and $β$ .
Sum of roots, $α$ + $β$ = $\frac{- 2 (p + q)}{2} = - (p + q)$
$\Rightarrow {(α + β)}^{2} = {(p + q)}^{2}$ .....(A)
$α β = \frac{p^{2} + q^{2}}{2}$
${(α - β)}^{2} = {(α + β)}^{2} - 4 α β = {(p + q)}^{2} - 4 (\frac{p^{2} + q^{2}}{2}) = {(p + q)}^{2} - 2 (p^{2} + q^{2}) = - {(p - q)}^{2} . . . . . (B)$
$A + B = {(p + q)}^{2} - {(p - q)}^{2} = (p + q - p + q) (p + q + p - q) [∵ x^{2} - y^{2} = (x - y) (x + y)] = 4 p q$
$A B = {(p + q)}^{2} [- {(p - q)}^{2}] = - (p^{2} + q^{2} + 2 p q) (p^{2} + q^{2} - 2 p q) = - [{(p^{2})}^{2} + {(q^{2})}^{2} - 2 p^{2} q^{2}] = - {(p^{2} - q^{2})}^{2}$
General form of quadratic equation is
$x^{2} - (A + B) x + A B = 0$
Putting the value of A and B we get
$x^{2} - 4 p q - {(p^{2} - q^{2})}^{2} = 0$ .

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Question 11:

Mukund possesses Rs 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.

Answer:

Let amount with Sagar be Rs x.
Amount with Mukund = Rs x + 50
The product of the amount they have is 15,000.
$x (x + 50) = 15000 \Rightarrow x^{2} + 50 x = 15000 \Rightarrow x^{2} + 50 x - 15000 = 0 \Rightarrow x^{2} + 150 x - 100 x - 15000 = 0 \Rightarrow x (x + 150) - 100 (x + 150) = 0 \Rightarrow (x - 100) (x + 150) = 0 \Rightarrow x = 100, - 150$
But amount cannot be negative so,
Amount with Sagar = Rs 100 and that with Mukund is Rs 150.

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Question 12:

The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.

Answer:

Let the smaller number be x.
Larger number be y.
The square of smaller number is twice the greater number.
$x^{2} = 2 y$
The difference between squares of two numbers is 120.
$y^{2} - x^{2} = 120 \Rightarrow y^{2} - 2 y - 120 = 0 \Rightarrow y^{2} - 12 y + 10 y - 120 = 0 \Rightarrow y (y - 12) + 10 (y - 12) = 0 \Rightarrow (y + 10) (y - 12) = 0 \Rightarrow y = 12, - 10$
So, the numbers are 12 as the larger number and the smaller number is
$x^{2} = 2 \times 12 = 24 \Rightarrow x = \pm \sqrt{24}$

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Question 13:

Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.

Answer:

Total oranges = 540
Let the total number of students be x.
Orange each student gets will be $\frac{540}{x}$
If 30 more students are there so, the total number of students will be x + 30
Number of oranges per person will be $\frac{540}{x + 30}$ .
If 30 students were more each would get 3 oranges less = $\frac{540}{x} - 3$
$\frac{540}{x + 30}$ = $\frac{540}{x} - 3$
$\Rightarrow 3 = 540 (\frac{1}{x} - \frac{1}{x + 30}) \Rightarrow 1 = 180 (\frac{x + 30 - x}{x (x + 30)}) \Rightarrow 1 = 180 (\frac{30}{x^{2} + 30 x}) \Rightarrow x^{2} + 30 x = 5400 \Rightarrow x^{2} + 30 x - 5400 = 0 \Rightarrow x^{2} + 90 x - 60 x - 5400 = 0 \Rightarrow x (x + 90) - 60 (x + 90) = 0 \Rightarrow (x - 60) (x + 90) = 0 \Rightarrow x = 60, - 90$
But number of students cannot be negative so, the number of students is 60.

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Question 14:

Mr. Dinesh owns an agricultural farm at village Talvel. The length of the farm is 10 meter more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is $\frac{1}{3}$ of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and of the pond.

Answer:

Breadth of farm = x
Length of farm = $2 x + 10$
Area of farm = $l b = x (2 x + 10) = 2 x^{2} + 10 x$
Side of pond = $\frac{x}{3}$
Area of pond = $s^{2} = {(\frac{x}{3})}^{2}$
Area of farm = 2(Area of pond)
$\Rightarrow 2 x^{2} + 10 x = 20 \times \frac{x^{2}}{9} \Rightarrow 18 x^{2} + 90 x = 20 x^{2} \Rightarrow 9 x^{2} + 45 x = 10 x^{2} \Rightarrow x^{2} - 45 x = 0 \Rightarrow x (x - 45) = 0 \Rightarrow x = 0, 45$
Length of farm = $2 \times 45 + 10 = 100 units$
Breadth of farm = 45 units
Side of pond = $\frac{45}{3} = 15$ units

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