question_answer

Solve the following quadratic equation for x:  \[{{x}^{2}}-2ax-(4{{b}^{2}}-{{a}^{2}})=0\]

Answer:

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

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

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

\[\therefore \,\,(x-a+2b)(x-a-2b)=0\]

\[\Rightarrow x=a-2b\] or \[a+2b\]

Hence,              \[x=a-2b\] or \[x=a+2b\]