# Solve the given quadratic equation by factorization method. 4x2−2(a2+b2)x+a2b2=0

amolent3u

## Answered question

2021-12-07

Solve the given quadratic equation by factorization method.

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

### Answer & Explanation

Step 1 Given

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

Step 2 Solving

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

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

$2x(2x-{a}^{2})-{b}^{2}(2x-{a}^{2})=0$

$(2x-{b}^{2})(2x-{a}^{2})=0$

$2x-{b}^{2}=0;2x-{a}^{2}=0$

$2x={b}^{2};2x={a}^{2}$

$x=\frac{{b}^{2}}{2};x=\frac{{a}^{2}}{2}$

$\therefore x=\frac{{a}^{2}}{2},\frac{{b}^{2}}{2}$

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