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

Question
Polynomial factorization
asked 2021-02-13
Solve the given quadratic equation by factorization method
\(\displaystyle{4}{x}^{{2}}-{2}{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}^{{2}}{b}^{{2}}={0}\)

Answers (1)

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
\(\displaystyle{4}{x}^{{2}}-{2}{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}^{{2}}{b}^{{2}}={0}\)
\(\displaystyle{\left({4}{x}^{{2}}-{2}{a}^{{2}}{x}\right)}-{\left({2}{b}^{{2}}{x}-{a}^{{2}}{b}^{{2}}\right)}={0}\)
\(\displaystyle{2}{x}{\left({2}{x}-{a}^{{2}}\right)}-{b}^{{2}}{\left({2}{x}-{a}^{{2}}\right)}={0}\)
\(\displaystyle{\left({2}{x}-{b}^{{2}}\right)}{\left({2}{x}-{a}^{{2}}\right)}={0}\)
\(\displaystyle{2}{x}-{b}^{{2}}={0},{2}{x}-{a}^{{2}}={0}\)
\(\displaystyle{2}{x}={b}^{{2}},{2}{x}={a}^{{2}}\)
\(\displaystyle{x}=\frac{{{b}^{{2}}}}{{2}},{x}=\frac{{{a}^{{2}}}}{{2}}\)
\(\displaystyle\therefore{x}=\frac{{{a}^{{2}}}}{{2}},\frac{{{b}^{{2}}}}{{2}}\)
0

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