Question

If $x-\frac{1}{x}=\sqrt{6}$, then $x^2+\frac{1}{x^2}$ is ________.

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (D)

$8$

Given, $\frac{x-1}{x}=6$

Multiplying and divide the above equation with $x-\frac{1}{x}$

Thus $\frac{x-\frac{1}{x}\times x-\frac{1}{x}}{x-\frac{1}{x}}=\sqrt{6}$

Using $(a-b{)}^2=a^2+b^2-2ab$

and substituting $x-\frac{1}{x}=\sqrt{6}$ in denominator, we get

$\frac{x^2+\frac{1}{x^2}-2x\frac{1}{x}}{\sqrt{6}}=\sqrt{6}$

$\Rightarrow x^2+\frac{1}{x^2}-2=\sqrt{6}\times \sqrt{6}$

$\Rightarrow x^2+\frac{1}{x^2}=6+2=8$