Question

If $x=2\sqrt{2}+\sqrt{7}$, identify the value of $\frac{1}{2}(x+\frac{1}{x})$.

• A. $2\sqrt{2}-\sqrt{7}$
• B. $\sqrt{7}$
• C. $2\sqrt{2}$
• D. $2\sqrt{2}+\sqrt{7}$

Answer 1

The correct option is C

$2\sqrt{2}$

$x=2\sqrt{2}+\sqrt{7}\frac{1}{2}(x+\frac{1}{x})=\frac{1}{2}(2\sqrt{2}+\sqrt{7}+\frac{1}{2\sqrt{2}+\sqrt{7}})=\frac{1}{2}\left(\right(\frac{2\sqrt{2}+\sqrt{7}{)}^2+1}{2\sqrt{2}+\sqrt{7}})=\frac{1}{2}(\frac{8+7+2\times 2\sqrt{2}\times \sqrt{7}+1}{2\sqrt{2}+\sqrt{7}})=\frac{1}{2}(\frac{16+4\sqrt{14}}{2\sqrt{2}+\sqrt{7}})=\frac{1}{2}(\frac{4(4+\sqrt{14})}{2\sqrt{2}+\sqrt{7}})=2(\frac{\sqrt{2}(2\sqrt{2}+\sqrt{7})}{2\sqrt{2}+\sqrt{7}})=2\sqrt{2}$