# Ncert Solutions For Class 10 Maths Ex 4.3

## Ncert Solutions For Class 10 Maths Chapter 4 Ex 4.3

Q.1 Find the roots of the equation 25x2 + 20x – 13 = 0 by method of completing the square.

Sol.

(5x)2 + 20x  = 13

(5x)2 + 2(5x.2)+ 22 = 13+22

(5x+2)2 – 17 = 0

Therefore  (5x+2)2(17)2=0$(5x+2)^{2}-(\sqrt{17})^{2}=0$

(5x+2+17)$(5x+2+\sqrt{17})$   (5x+217)=0$(5x+2-\sqrt{17})=0$

Therefore     x=2+175$x=-\frac{2+\sqrt{17}}{5}$   or   x=2+175$x=\frac{-2+\sqrt{17}}{5}$

Q.2 Find the roots of the equation 9x2 + 12x – 5 = 0 by method of completing the square.

Sol.

9x2 + 12x = 5

(3x)2 + 2(3x.2) = 5

(3x)2 + 2(3x.2) + 22 = 5 + 22

(3x+2)2 = 9

(3x+2)2 – 32 = 0

(3x + 2 + 3) (3x + 2 – 3) = 0

(3x + 5) (3x – 1) = 0

Therefore   x=53$x=\frac{-5}{3}$   or   x=13$x =\frac{1}{3}$

Q.3 Find the roots of the equation 3x2 – 16x + 5 = 0 by method of completing the square.

Sol.

To solve the above equation either divide the entire equation by 3 to make 3x2 a perfect square but it will be much easier to solve the equation if we multiply the above equation by 3 to make 3x2 a perfect square.

Therefore,           9x2 – 48x = -15

(3x)2 – 2(3x.8) =-15

(3x)2 – 2( 3x.8 ) + 82 = -15 + 82

(3x – 8)2 – 49 = 0

(3x-8)2 – 72 = 0

(3x-8+7)(3x-8-7)=0

(3x-1)(3x-15)=0

Therefore       x=13$x=\frac{1}{3}$   or   x=5$x=5$

Q.4 Find the roots of the equation 2x2 + x – 4 = 0 by method of completing the square.

Sol.

Now, for making 2x2 a perfect square, multiplying the above equation by 2

Therefore,

4x2 + 2x = 8

Now,

(2x)2+2(2x.12)=8$(2x)^{2} + 2(2x.\frac{1}{2})=8$

(2x)2+2(2x.12)+14=8+14$(2x)^{2}+2(2x.\frac{1}{2})+\frac{1}{4}= 8+\frac{1}{4}$

(2x+12)2(334)2=0$(2x+\frac{1}{2})^{2}-(\sqrt{\frac{33}{4}})^{2}=0$

(2x+12+332)(2x+12332)=0$(2x+\frac{1}{2}+\frac{\sqrt{33}}{2})\;\;(2x+\frac{1}{2}-\frac{\sqrt{33}}{2})=0$

Therefore   x=1+334$x=-\frac{1+\sqrt{33}}{4}$  or  x=1+334$x=\frac{-1+\sqrt{33}}{4}$

ax2 + bx+ c = 0 with a ≠ 0

The roots of the following quadratic equation are given by:

x=b+b24ac2a$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$  and  x=bb24ac2a$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

(or)      x=b±b24ac2a$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

This is known as quadratic formulae for finding roots of quadratic equation.

Nature of roots:

The term (b2 – 4ac) is known as discriminant and it determines the nature of roots of the quadratic equation.

If b2 – 4ac > 0, then the equation will have two distant real roots.

If b2 – 4ac = 0, then the equation will have two equal real roots.

If b2 – 4ac < 0, then the equation will have unreal roots.