Question

If $x=\frac{2\sqrt{5}}{\sqrt{3}+\sqrt{5}}$, then what is the value of $\frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$

• A. $\sqrt{5}$
• B. $\sqrt{3}$
• C. $\sqrt{15}$
• D. $2$

Answer 1

Answer: (D)

$2$

$x=\frac{2\sqrt{5}}{\sqrt{3}+\sqrt{5}}\Rightarrow \frac{x}{\sqrt{3}}=\frac{2\sqrt{5}}{\sqrt{3}+\sqrt{5}}$
and $\frac{x}{\sqrt{5}}=\frac{2\sqrt{3}}{\sqrt{3}+\sqrt{5}}$
Applying components and dividendo, we get
$\frac{x+\sqrt{5}}{x-\sqrt{5}}=-(7+2\sqrt{15})$
and
$\frac{x+\sqrt{3}}{x-\sqrt{3}}=9+2\sqrt{15}$
$\Rightarrow \frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=2$