Question

Solve the given quadratic equation by factorization method
$4x^2-4ax+(a^2-b^2)=0$

Answer 1

We have

$4x^2-4ax+(a^2-b^2)=0$

Here, Constant term $=(a^2-b^2)=(a-b)(a+b)$

and coefficient of the middle term $=-4a$

Also. coefficient of the middle term $-4a=-[2a+2b+2a-2b]=-\left[2\right(a+b)+2(a-b\left)\right]$

$\therefore 4x^2-4ax+(a^2-b^2)=0$

$\Rightarrow 4x^2-\left[2\right(a+b)+2(a-b\left)\right]x+(a+b)(a-b)=0$

$\Rightarrow 4x^2-2(a+b)x-2(a-b)x+(a+b)(a-b)=0$

$\Rightarrow [4x^2-2(a+b\left)x\right]-\left[2\right(a-b)x-(a+b\left)\right(a-b\left)\right]=0$

$\Rightarrow 2x[2x-(a+b\left)\right]-(a-b)[2x-(a+b\left)\right]=0$

$\Rightarrow [2x-(a+b\left)\right][2x-(a-b\left)\right]=0$

$\Rightarrow [2x-(a+b\left)\right]=0or[2x-(a-b\left)\right]=0$

$\Rightarrow 2x=a+bor2x=a-b$

$\Rightarrow x=\frac{a+b}{2}orx=\frac{a-b}{2}$