Question

If $x=2+\sqrt{3}$ and $xy=1$, then $\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}=?$

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. 1
• D. None of these

Answer 1

The correct option is A $\sqrt{2}$

Given, $x=2+\sqrt{3}$.....(1).

And $xy=1$ then,

$y=\frac{1}{x}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$......(2).

Now from (1) we get,

$x=\frac{1}{2}(4+2\sqrt{3})=\frac{1}{2}(\sqrt{3}+1{)}^2$.

Then, $\sqrt{x}=\frac{1}{\sqrt{2}}(\sqrt{3}+1)$.

In the same way we can find $\sqrt{y}=\frac{1}{\sqrt{2}}(\sqrt{3}-1)$.

Now,

$\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}$

$=\frac{\frac{(\sqrt{3}+1{)}^2}{2}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}}$$+\frac{\frac{(\sqrt{3}-1{)}^2}{2}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}}$

$=\frac{\frac{(\sqrt{3}+1{)}^2}{\sqrt{2}}}{\sqrt{3}+3}$$+\frac{\frac{(\sqrt{3}-1{)}^2}{\sqrt{2}}}{3-\sqrt{3}}$

$=\frac{(\sqrt{3}+1)}{\sqrt{2}\times \sqrt{3}}$$+\frac{(\sqrt{3}-1)}{\sqrt{2}\times \sqrt{3}}$

$=\frac{1}{\sqrt{6}}\times 2\sqrt{3}$

$=\sqrt{2}$.