# Frame Quadratic Equations & Find Roots

Exercise 4.3 Q3. NCERT Solutions For Class 10 Maths

#### Question: Find the roots of the following equations:

∴ x = $$frac{-$-3$士 {$sqrt{13}}}{2}) = $\frac{3 士 {\sqrt{13}}}{2}$ x = $\frac{3 + {\sqrt{13}}}{2}$ and $\frac{3 - {\sqrt{13}}}{2}$ ##### ii. $\frac{1}{x + 4}$ - $\frac{1}{x - 7}$ = $\frac{11}{30}$, x ≠ -4, 7 Because x ≠ -4, 7 multiply the equation by$x + 4) (x - 7)
(x - 7) - (x + 4) = $$frac{11}{30}$$x + 4) (x - 7)
-11 = $$frac{11}{30}$$x + 4) (x - 7)
Multiply the equation by $$frac{30}{11}$ -30 =$x + 4) (x - 7)
= x2 - 7x + 4x - 28 + 30 = 0
x2 - 3x + 2 = 0

##### Method 1: Factorize the quadratic equation to find the values of x

x2 - 2x - x + 2 = x(x - 2) - 1 (x - 2) = 0
(x - 2 )(x - 1) = 0
x = 2 and x = 1

##### Method 2: Use the quadratic formula to find the values of x
The roots of ax2 + bx + c = 0 are $$frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a}$ In this equation, a = 1, b = -3 and c = 2 b2 – 4ac =$-3)2 – 4 * 1 * 2 = 9 – 8 = 1 > 0
Therefore, x = -$$frac{-$-3$士{$sqrt1}}{2}$ = $\frac{3 士 1}{2}$
x= 2 and x = 1