Question

The value of $\frac{1}{x^2}+\frac{1}{y^2}$, where $x=2+\sqrt{3}$ and $y=2-\sqrt{3}$, is?

• A. $12$
• B. $16$
• C. $14$
• D. $10$

Answer 1

Answer: (C)

$14$

$x=2+\sqrt{3}$, $y=2-\sqrt{3}$
$\frac{1}{x}=\frac{1}{2+\sqrt{3}}$ $\frac{1}{y}=\frac{1}{2-\sqrt{3}}$
By Rationalizing
$\frac{1}{x}=\frac{1}{(2+\sqrt{3})}\frac{(2-\sqrt{3})}{(2-\sqrt{3})},\frac{1}{y}=\frac{(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$
$\frac{1}{x}=2-\sqrt{3}$ , $\frac{1}{y}=2+\sqrt{3}$
$\frac{1}{x^2}={(2-\sqrt{3})}^2$ $\frac{1}{y^2}={(2+\sqrt{3})}^2$
$\frac{1}{x^2}+\frac{1}{y^2}=2^2+(\sqrt{3}{)}^2-4\sqrt{3}+2^2+{\left(\sqrt{3}\right)}^2+4\sqrt{3}$
$\frac{1}{x^2}+\frac{1}{y^2}=4+4+3+3=14$.