If $\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$, then the value of $x^2+\frac{1}{x^2}$ is:
Answer & explanation
Correct answer: option 3
If $K-\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 + 2
If $\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$,
then the value of $x^2+\frac{1}{x^2}$
= $x^2+\frac{1}{x^2}$ = $\sqrt{(\sqrt{7})^2 + 2}$ = 3