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

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

2021-02-14
Step 1
Given
$$\displaystyle{4}{x}^{{2}}-{2}{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}^{{2}}{b}^{{2}}={0}$$
Step 2 Solving
$$\displaystyle{4}{x}^{{2}}-{2}{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}^{{2}}{b}^{{2}}={0}$$
$$\displaystyle{\left({4}{x}^{{2}}-{2}{a}^{{2}}{x}\right)}-{\left({2}{b}^{{2}}{x}-{a}^{{2}}{b}^{{2}}\right)}={0}$$
$$\displaystyle{2}{x}{\left({2}{x}-{a}^{{2}}\right)}-{b}^{{2}}{\left({2}{x}-{a}^{{2}}\right)}={0}$$
$$\displaystyle{\left({2}{x}-{b}^{{2}}\right)}{\left({2}{x}-{a}^{{2}}\right)}={0}$$
$$\displaystyle{2}{x}-{b}^{{2}}={0},{2}{x}-{a}^{{2}}={0}$$
$$\displaystyle{2}{x}={b}^{{2}},{2}{x}={a}^{{2}}$$
$$\displaystyle{x}=\frac{{{b}^{{2}}}}{{2}},{x}=\frac{{{a}^{{2}}}}{{2}}$$
$$\displaystyle\therefore{x}=\frac{{{a}^{{2}}}}{{2}},\frac{{{b}^{{2}}}}{{2}}$$

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