Solve this
Question:

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

 

Solution:

Given :

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

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

$\Rightarrow 2 x\left(2 x-a^{2}\right)-b^{2}\left(2 x-a^{2}\right)=0$

$\Rightarrow\left(2 x-b^{2}\right)\left(2 x-a^{2}\right)=0$

$\Rightarrow 2 x-b^{2}=0$ or $2 x-a^{2}=0$

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

Hence, the roots of the equation are $\frac{b^{2}}{2}$ and $\frac{a^{2}}{2}$.

 

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