question_answer

Solve the following quadratic equation for x:

\[4{{x}^{2}}-4{{a}^{2}}x+({{a}^{4}}-{{b}^{4}})=0\].

Answer:

We have \[4{{x}^{2}}-4{{a}^{2}}x+({{a}^{4}}-{{b}^{4}})=0\]

\[(4{{x}^{2}}-4{{a}^{2}}x+{{a}^{4}})-{{b}^{4}}=0\]

\[{{(2x-{{a}^{2}})}^{2}}-{{({{b}^{2}})}^{2}}=0\]

\[\therefore (2x-{{a}^{2}}+{{b}^{2}})(2x-{{a}^{2}}-{{b}^{2}})=0\]

\[x=\frac{{{a}^{2}}-{{b}^{2}}}{2}\] or \[\frac{{{a}^{2}}+{{b}^{2}}}{2}\]