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 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.