Question

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

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $1$
• D. $2$

Answer 1

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

Given $x=2+\sqrt{3}$ and $xy=1$

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

Let $\sqrt{x}=\sqrt{a}+\sqrt{b}$

then $x=a+b+2\sqrt{ab}$ by squaring both sides

We have $x=2+\sqrt{3}=a+b+2\sqrt{ab}$

$\Rightarrow a+b=2,\sqrt{ab}=\frac{\sqrt{3}}{2}$ or $ab=\frac{3}{4}$

$\Rightarrow {(a-b)}^2={(a+b)}^2-4ab=4-4\times \frac{3}{4}=4-3=1$

$\Rightarrow a-b=1$

$\Rightarrow a+b=2,a-b=1$

$\Rightarrow 2a=3$

$\Rightarrow a=\frac{3}{2}$

Put $a=\frac{3}{2}$ in $a+b=2$

$b=2-a=2-\frac{3}{2}=\frac{4-3}{2}=\frac{1}{2}$

$\therefore a=\frac{3}{2},b=\frac{1}{2}$

So,$\sqrt{x}=\frac{\sqrt{3}+1}{\sqrt{2}}$

$\sqrt{y}=\frac{1}{\sqrt{x}}=\frac{\sqrt{2}}{\sqrt{3}+1}$

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

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

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

Now substitute in the given expression:

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

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

$=\frac{\sqrt{2}(2+\sqrt{3})}{3+\sqrt{3}}\times \frac{3-\sqrt{3}}{3-\sqrt{3}}=\frac{\sqrt{2}(3+\sqrt{3})}{6}$

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

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

$=\frac{\sqrt{2}(2-\sqrt{3})}{3-\sqrt{3}}\times \frac{3+\sqrt{3}}{3+\sqrt{3}}=\frac{\sqrt{2}(3-\sqrt{3})}{6}$

Now,$\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}=\frac{\sqrt{2}(3+\sqrt{3})}{6}+\frac{\sqrt{2}(3-\sqrt{3})}{6}$

$=\frac{\sqrt{2}(3+\sqrt{3}+3-\sqrt{3})}{6}$

$=\frac{6\sqrt{2}}{6}=\sqrt{2}$