Question

If $x=\sqrt{7}+\sqrt{6}$, then ${(x+\frac{1}{x})}^2$=

• A. $2(\sqrt{7}+\sqrt{6})$
• B. $2\sqrt{6}$
• C. $28$
• D. $2\sqrt{7}$

Answer 1

The correct option is C $28$

$x=\sqrt{7}+\sqrt{6},\frac{1}{x}=\frac{1}{\sqrt{7}+\sqrt{6}}\times \frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}-\sqrt{6}}=\frac{\sqrt{7}-\sqrt{6}}{\left(\sqrt{7}{)}^2\right(\sqrt{6}{)}^2}=\sqrt{7}-\sqrt{6}$

$\Rightarrow x+\frac{1}{x}=\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6}=2\sqrt{7}$

$\Rightarrow (x+\frac{1}{x}{)}^2=28\Rightarrow (c)$