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

Solve the given quadratic equation by factorization method
$4{x}^{2}-2\left({a}^{2}+{b}^{2}\right)x+{a}^{2}{b}^{2}=0$
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Nathanael Webber
Step 1
Given
$4{x}^{2}-2\left({a}^{2}+{b}^{2}\right)x+{a}^{2}{b}^{2}=0$
Step 2 Solving
$4{x}^{2}-2\left({a}^{2}+{b}^{2}\right)x+{a}^{2}{b}^{2}=0$
$\left(4{x}^{2}-2{a}^{2}x\right)-\left(2{b}^{2}x-{a}^{2}{b}^{2}\right)=0$
$2x\left(2x-{a}^{2}\right)-{b}^{2}\left(2x-{a}^{2}\right)=0$
$\left(2x-{b}^{2}\right)\left(2x-{a}^{2}\right)=0$
$2x-{b}^{2}=0,2x-{a}^{2}=0$
$2x={b}^{2},2x={a}^{2}$
$x=\frac{{b}^{2}}{2},x=\frac{{a}^{2}}{2}$
$\therefore x=\frac{{a}^{2}}{2},\frac{{b}^{2}}{2}$
Jeffrey Jordon