amolent3u

2021-12-07

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

servidopolisxv

Step 1 Given
$4{x}^{2}-2\left({a}^{2}+{b}^{2}\right)x+{a}^{2}{b}^{2}=0$
Step 2 Solving
$4{x}^{2}-2{a}^{2}x-2{b}^{2}x+{a}^{2}{b}^{2}=0$
$\left(4{x}^{2}-2{a}^{2}x\right)-\left(2{b}^{2}x-{a}^{2}{b}^{2}\right)=0$
$2x\left(2x-{a}^{2}\right)-{b}^{2}\left(2x-{a}^{2}\right)=0$
$\left(2x-{b}^{2}\right)\left(2x-{a}^{2}\right)=0$
$2x-{b}^{2}=0;2x-{a}^{2}=0$
$2x={b}^{2};2x={a}^{2}$
$x=\frac{{b}^{2}}{2};x=\frac{{a}^{2}}{2}$
$\therefore x=\frac{{a}^{2}}{2},\frac{{b}^{2}}{2}$

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