Question
Solve the following quadratic equation: $4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$

Answer

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

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