Question

If $x=8+3\sqrt{7}\mathrm{and}xy=1$, then $\frac{1}{x^2}+\frac{1}{y^2}$ is:

• A. $254$
• B. $192$
• C. $292$
• D. $66$
• E. $62$

Answer 1

The correct option is A

$254$

Explanation for the correct option.

Step 1: Find the value of $\frac{1}{x}$.

$$\begin{gathered} \begin{array}{rcl}x& =& 8+3\sqrt{7}\\ & \Rightarrow & \frac{1}{x}=\frac{1}{8+3\sqrt{7}}\\ & =& \frac{1}{8+3\sqrt{7}}\times \frac{8-3\sqrt{7}}{8-3\sqrt{7}}\\ & =& \frac{8-3\sqrt{7}}{8^2-{\left(3\sqrt{7}\right)}^2}\end{array} \\ =\frac{8-3\sqrt{7}}{64-63} \\ =8-3\sqrt{7} \end{gathered}$$

Step 2: Find the value of $\frac{1}{y}$.

$\begin{array}{rcl}xy& =& 1\\ \Rightarrow x& =& \frac{1}{y}\\ & \Rightarrow & \frac{1}{y}=8+3\sqrt{7}\end{array}$

Step 3: Find the value of $\frac{1}{x^2}+\frac{1}{y^2}$.

$\begin{array}{rcl}\frac{1}{x^2}+\frac{1}{y^2}& =& {\left(\frac{1}{x}\right)}^2+{\left(x\right)}^2\\ & =& {\left(8-3\sqrt{7}\right)}^2+{\left(8+3\sqrt{7}\right)}^2\\ & =& \left(64-48\sqrt{7}+63\right)+\left(64+48\sqrt{7}+63\right)\\ & =& 254\end{array}$

Hence, option A is correct.