Question

If $x=7+4\sqrt{3}$ and $xy=1$, what is the value of $\frac{1}{x^2}+\frac{1}{y^2}$?

• A. $64$
• B. $134$
• C. $194$
• D. $\frac{1}{49}$

Answer 1

Answer: (C)

$194$

$x=7+4\sqrt{3}$ & $xy=1$

$\Rightarrow y=\frac{1}{7+4\sqrt{3}}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$

$\frac{1}{x^2}+\frac{1}{y^2}=(\frac{1}{x}+\frac{1}{y}{)}^2-2\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}=(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}{)}^2-2$

$=(\frac{14}{49-48}{)}^2-2=196-2$

$=194$