question_answer

If  \[x-\frac{1}{x}=7,\] then find the value \[{{x}^{2}}+\frac{1}{{{x}^{2}}}\]

Answer:

We know that

\[{{(ab)}^{2}}={{a}^{2}}\text{ }2ab+{{b}^{2}}\]

\[\therefore\]

\[{{\left( x-\frac{1}{x} \right)}^{2}}={{(x)}^{2}}-2(x)\left( \frac{1}{x} \right)+{{\left( \frac{1}{x} \right)}^{2}}\]

\[{{\left( x-\frac{1}{x} \right)}^{2}}={{x}^{2}}-2+\frac{1}{{{x}^{2}}}\]

\[{{(7)}^{2}}={{x}^{2}}-2+\frac{1}{{{x}^{2}}}\]

\[49+2={{x}^{2}}+\frac{1}{{{x}^{2}}}\]

\[{{x}^{2}}+\frac{1}{{{x}^{2}}}=51\]