RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.4

RD Sharma Solutions Class 10 Chapter 8 Exercise 8.4

RD Sharma Class 10 Solutions Chapter 8 Ex 8.4 PDF Free Download

Exercise 8.4

By using the method of completing the square, find the roots of the following quadratic equations: (1)  $x^{2} – 4 \sqrt{2x} + 6 = 0$

Soln:

$x^{2} – 4 \sqrt{2x} + 6 = 0$

i.e. $x^{2} – 2 \times x \times 2 \sqrt{2} + (2 \sqrt{2})^{2} – (2\sqrt{2})^{2} + 6 = 0$

$(x-2\sqrt{2})^{2} = (2\sqrt{2})^{2} – 6$

=$(x-2\sqrt{2})^{2} = (4\times 2) – 6$

= $(x-2\sqrt{2})^{2} = 8-6$

= $(x-2\sqrt{2})^{2} = 2$

= $(x-2\sqrt{2})$ = $\pm\sqrt{2}$

= $(x-2\sqrt{2})$ = $\sqrt{2}$  or  $(x-2\sqrt{2})$ =  –  $\sqrt{2}$

$x = \sqrt{2} + 2\sqrt{2}$  or    $x = – \sqrt{2} + 2\sqrt{2}$

= $x = 3\sqrt{2}$  or  $x = \sqrt{2}$

So, the roots for the given equation are :

$x = 3\sqrt{2}$  or  $x = \sqrt{2}$

(2)  $2x^{2} – 7x + 3 = 0$

Soln:

$2x^{2} – 7x + 3 = 0$

$2(x^{2} – \frac{7x}{2} + \frac{3}{2}) = 0$

$x^{2} – 2 \times \frac{7}{2} \times \frac{1}{2} \times x + \frac{3}{2} = 0$

$x^{2} – 2\times \frac{7}{4} \times x + (\frac{7}{4})^{2} – (\frac{7}{4})^{2} + \frac{3}{2} = 0$

$x^{2} – 2\times \frac{7}{4}\times x + (\frac{7}{4})^{2} – (\frac{49}{16}) + \frac{3}{2} = 0$

$(x – \frac{7}{4})^{2} – \frac{49}{16} + \frac{3}{2} = 0$

$(x – \frac{7}{4})^{2}=\frac{49}{16} – \frac{3}{2}$

$(x – \frac{7}{4})^{2}=\frac{49-26}{16}$

$(x – \frac{7}{4})^{2}= \frac{25}{16}$

$(x – \frac{7}{4})^{2}= (\frac{5}{4})^{2}$

$x-\frac{7}{4} = \pm \frac{5}{4}$

$x – \frac{7}{4} = \frac{5}{4}$  or  $x – \frac{7}{4} = – \frac{5}{4}$

$x = \frac{7}{4} + \frac{5}{4}$  or  $x = \frac{7}{4} – \frac{5}{4}$

$x = \frac{12}{4}$  or  $x = \frac{2}{4}$

x = 3  or  x = 1/2

(3)  $3x^{2}+11x+10 = 0$

Soln:  $3x^{2} + 11x + 10 = 0$

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

$x^{2} + 2\times \frac{1}{2}\times \frac{11x}{3} + \frac{10}{3} = 0$

$x^{2} + 2\times \frac{11x}{6} + (\frac{11}{6})^{2} – (\frac{11}{6})^{2} + \frac{10}{3} = 0$

$(x + \frac {11}{6})^{2} = (\frac{11}{6})^{2} – \frac{10}{3}$

$(x + \frac {11}{6})^{2} = \frac{121}{36} – \frac{10}{3}$

$(x + \frac {11}{6})^{2} = \frac{121 – 120}{36}$

$(x + \frac {11}{6})^{2} = \frac{1}{36}$

$(x + \frac {11}{6})^{2} = (\frac{1}{6})^{2}$

$x + \frac {11}{6} = \pm \frac{1}{6}$

$x + \frac {11}{6} = \frac{1}{6}$  or  $x + \frac{11}{6} = – \frac{1}{6}$

$x = \frac {1}{6} – \frac{11}{6}$  or  $x=\frac{-1}{6} – \frac{11}{6}$

$x = \frac {-10}{6}$  or  $x = \frac{-12}{6}$  = – 2

x = – 5/3   or  x = – 2

(4)  $2x^{2} + x – 4= 0$

Soln:  $2x^{2} + x – 4= 0$

$2(x^{2} + \frac{x}{2} – \frac{4}{2}) = 0$

$x^{2} + 2\times \frac{1}{2}\times \frac{1}{2}\times x – 2=0$

$x^{2} + 2\times \frac{1}{4}\times x + (\frac{1}{4})^{2} – (\frac{1}{4})^{2} – 2=0$

$(x + \frac{1}{4})^{2} = (\frac{1}{4})^{2} +2$

$(x +\frac{1}{4})^{2}$ = $(\frac{1}{4})^{2} + 2$

$(x + \frac{1}{4})^{2}$ = $\frac{1}{16} +2$

$(x + \frac{1}{4})^{2}$ = $\frac{1 + 2\times 16}{16}$

$(x +\frac{1}{4})^{2}$ = $\frac{1 + 32}{16}$

$(x + \frac{1}{4})^{2}$ = $\frac{33}{16}$

$(x + \frac{1}{4})$ = $\pm \sqrt{\frac{33}{16}}$

$(x + \frac{1}{4})$ = $\sqrt{\frac{33}{16}}$   or  $(x+\frac{1}{4})$ = – $\sqrt{\frac{33}{16}}$

$x=\frac{\sqrt{33}}{4} – \frac{1}{4}$  or  $x= – \frac{\sqrt{33}}{4} – \frac{1}{4}$

$x=\frac{\sqrt{33} – 1}{4}$  or  $x=\frac{ – \sqrt{33}-1}{4}$

So, $x = \frac{\sqrt{33} – 1}{4}$  or  $x= \frac{ – \sqrt{33} – 1}{4}$

Are the two roots of the given equation.

(5)  $2x^{2} + x + 4=0$

Soln:  $2x^{2} + x + 4 = 0$

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

$x^{2} + 2\times \frac{1}{2}\times \frac{1}{2}\times x + 2=0$

$x^{2} + 2 \times \frac{1}{4} \times x + (\frac{1}{4})^{2} – (\frac{1}{4})^{2} + 2=0$

$x^{2} + 2 \times \frac{1}{4} \times x + (\frac{1}{4})^{2} = (\frac{1}{4})^{2} – 2$

$(x + \frac{1}{4})^{2} = \frac{1}{16} – 2$

$(x + \frac{1}{4})^{2}$ = $\frac{1-32}{16}$

$(x + \frac{1}{4})^{2}$ = $\frac{-31}{16}$

$(x + \frac{1}{4}$ = $\pm \sqrt{\frac{- 31}{16}}$

$(x + \frac{1}{4}$ = $\frac{\sqrt{- 31}}{4}$  or

$(x + \frac{1}{4}$ = $\frac{ – \sqrt{ – 31}}{4}$

$x = \frac{\sqrt{-31}-1}{4}$  or  $x = \frac{ – \sqrt{-31}-1}{4}$

Since,  $\sqrt{-31}$ is not a real number,

Therefore, the equation doesn’t have real roots.

(6)  $4x^{2} + 4\sqrt{3}x + 3 = 0$

Soln:  $4x^{2} + 4\sqrt{3}x + 3 = 0$

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

$x^{2} + 2 \times \frac{1}{2} \times \sqrt{3} \times x + \frac{3}{4} = 0$

$x^{2} + 2 \times \frac{\sqrt{3}}{2} \times x + (\frac{\sqrt{3}}{2})^{2} – (\frac{\sqrt{3}}{2})^{2} + \frac{3}{4} = 0$

$(x + \frac{\sqrt{3}}{2})^{2} – \frac{3}{4} + \frac{3}{4} = 0$

$(x + \frac{\sqrt{3}}{2})^{2}$ = 0

$(x + \frac{\sqrt{3}}{2})$ = 0  and  $(x + \frac{\sqrt{3}}{2})$ = 0

$x = \frac{-\sqrt{3}}{2}$  and $x = \frac{-\sqrt{3}}{2}$

Therefore, $x = \frac{-\sqrt{3}}{2}$  and $x = \frac{-\sqrt{3}}{2}$

Are the real roots of the given equation.

(7)  $\sqrt{2}x^{2} – 3x – 2\sqrt{2} = 0$

Soln:   $\sqrt{2}x^{2} – 3x – 2\sqrt{2} = 0$

$x^{2} – \frac{3x}{\sqrt{2}} – \frac{2\sqrt{2}}{\sqrt{2}} = 0$

$x^{2} – \frac{3x}{\sqrt{2}} – 2 = 0$

$x^{2} – 2 \times \frac{1}{2} \times \frac{3x}{\sqrt{2}} – 2 = 0$

$x^{2} – 2 \times \frac{3x}{2\sqrt{2}} + (\frac{3}{2\sqrt{2}})^{2} – (\frac{3}{2\sqrt{2}})^{2} – 2 = 0$

$(x-\frac{3}{2\sqrt{2}})^{2} = \frac{9}{8} + 2$

$(x-\frac{3}{2\sqrt{2}})^{2} = \frac{9+16}{8}$

$(x-\frac{3}{2\sqrt{2}})^{2} = \frac{25}{8}$

(8)  $\sqrt{3}x^{2} + 10x + 7\sqrt{3} = 0$

Soln:  $\sqrt{3}x^{2} + 10x + 7\sqrt{3} = 0$

$x^{2} + \frac{10x}{\sqrt{3}} + \frac{7\sqrt{3}}{\sqrt{3}} = 0$

$x^{2} + 2\times \frac{1}{2}\times \frac{10x}{\sqrt{3}} + 7 = 0$

$(x + \frac{5}{\sqrt{3}})^{2} = \frac{25}{3} – 7$

$(x + \frac{5}{\sqrt{3}})^{2} = \frac{25 – 21}{3}$

$(x + \frac{5}{\sqrt{3}})^{2} = \frac{4}{3}$

$x + \frac{5}{\sqrt{3}} = \pm \sqrt{\frac{4}{3}}$

$x + \frac{5}{\sqrt{3}} = + \frac{2}{\sqrt{3}} or x + \frac{5}{\sqrt{3}} = – \frac{2}{\sqrt{3}}$

$x = \frac{-3}{\sqrt{3}} or x = \frac{-7}{\sqrt{3}}$

$x = -\sqrt{3} and x = \frac{-7}{\sqrt{3}}$

(9)  $x^{2} – (\sqrt{2} + 1)x + \sqrt{2} = 0$

Soln: $x^{2} – (\sqrt{2} + 1)x + \sqrt{2} = 0$

$x^{2} – 2\times \frac{1}{2}(\sqrt{2} + 1)x + \sqrt{2} = 0$

$x^{2} – 2\times \frac{\sqrt{2}+1}{2}x + (\frac{\sqrt{2}+1}{2})^{2} – (\frac{\sqrt{2}+1}{2})^{2} + \sqrt{2} = 0$

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

$(x – \frac{\sqrt{2}+1}{2})^{2}$ = $\frac{(2 + 2\sqrt{2} + 1)}{4} – \sqrt{2}$

$(x – \frac{\sqrt{2}+1}{2})^{2}$ = $\frac{(3 + 2\sqrt{2} – 4\sqrt{2})}{4}$

$(x – \frac{\sqrt{2}+1}{2})^{2}$ = $\frac{(3 – 2\sqrt{2})}{4}$

$(x – \frac{\sqrt{2}+1}{2})^{2}$ =  $\frac{(2 + 1 – 2\sqrt{2})}{4}$

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

$(x – \frac{\sqrt{2}+1}{2})^{2}$ = $\frac{(\sqrt{2}-1)^{2}}{(2)^{2}}$

$(x – \frac{\sqrt{2}+1}{2})^{2}$ = $(\frac{\sqrt{2}-1}{2})^{2}$

$x – \frac{\sqrt{2}+1}{2}$ = $\pm$ $(\frac{\sqrt{2}-1}{2})$

$x – \frac{\sqrt{2}+1}{2}$ =  $(\frac{\sqrt{2}-1}{2})$

Or  $x – \frac{\sqrt{2}+1}{2}$ =  $(\frac{-\sqrt{2}-1}{2})$

$x = \frac{\sqrt{2}+1}{2}$ +  $(\frac{\sqrt{2}-1}{2})$

Or  $x = \frac{\sqrt{2}+1}{2}$ +  $(\frac{-\sqrt{2}-1}{2})$

$x = \frac{2\sqrt{2}}{2} or x = \frac{2}{2}$

$x = \sqrt{2}$  or  x = 1.

(10)  x2 – 4ax + 4a2 – b2 = 0

Soln:   x2 – 4ax + 4a2 – b2 = 0

x2 – 2(2a).x + (2a)2 – b2 = 0

(x – 2a)2 = b2

x – 2a = $\pm$ b

x – 2a = b or x – 2a = – b

x = 2a + b or  x = 2a – b

Therefore, x = 2a + b or  x = 2a – b are the two roots of the given equation.

Practise This Question

Observe the pattern given below

Then term at nth position will be given by expression