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 \[\frac{{{a}^{2}}+{{b}^{2}}}{2}\]
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