Question

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

Answer 1

We have
$4x^2-2(a^2+b^2)x+a^2{b}^2=0$

Here constant term$=a^2{b}^2=a^2{\times b}^2$

and coefficient of middle term $=-2(a^2+b^2)$

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

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

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

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