Step 1

Given

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

Step 2 Solving

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

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

\(\displaystyle{2}{x}{\left({2}{x}-{a}^{{2}}\right)}-{b}^{{2}}{\left({2}{x}-{a}^{{2}}\right)}={0}\)

\(\displaystyle{\left({2}{x}-{b}^{{2}}\right)}{\left({2}{x}-{a}^{{2}}\right)}={0}\)

\(\displaystyle{2}{x}-{b}^{{2}}={0},{2}{x}-{a}^{{2}}={0}\)

\(\displaystyle{2}{x}={b}^{{2}},{2}{x}={a}^{{2}}\)

\(\displaystyle{x}=\frac{{{b}^{{2}}}}{{2}},{x}=\frac{{{a}^{{2}}}}{{2}}\)

\(\displaystyle\therefore{x}=\frac{{{a}^{{2}}}}{{2}},\frac{{{b}^{{2}}}}{{2}}\)

Given

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

Step 2 Solving

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

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

\(\displaystyle{2}{x}{\left({2}{x}-{a}^{{2}}\right)}-{b}^{{2}}{\left({2}{x}-{a}^{{2}}\right)}={0}\)

\(\displaystyle{\left({2}{x}-{b}^{{2}}\right)}{\left({2}{x}-{a}^{{2}}\right)}={0}\)

\(\displaystyle{2}{x}-{b}^{{2}}={0},{2}{x}-{a}^{{2}}={0}\)

\(\displaystyle{2}{x}={b}^{{2}},{2}{x}={a}^{{2}}\)

\(\displaystyle{x}=\frac{{{b}^{{2}}}}{{2}},{x}=\frac{{{a}^{{2}}}}{{2}}\)

\(\displaystyle\therefore{x}=\frac{{{a}^{{2}}}}{{2}},\frac{{{b}^{{2}}}}{{2}}\)