Question

If $N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$, then $N$ equals_________

• A. $1$
• B. $2\sqrt{2}-1$
• C. $\frac{\sqrt{5}}{2}$
• D. $\frac{2}{\sqrt{\sqrt{5}}+1}$

Answer 1

Answer: (A)

$1$

$N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$

Consider,

$A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}$

Squaring both sides

$\begin{array}{c}{A}^2=\sqrt{5}+2+\sqrt{5}-2+2\left(\sqrt{\sqrt{5}+2}\right)\left(\sqrt{\sqrt{5}-2}\right)\\ \Rightarrow A^2=2\sqrt{5}+2\sqrt{5-4}\\ \Rightarrow A^2=2\sqrt{5}+2\\ \Rightarrow A^2=2\left(\sqrt{5}+1\right)\end{array}$

So,

$A=\pm \sqrt{2}\sqrt{\sqrt{5}+1}$

Now Consider

$\begin{array}{c}\sqrt{3-2\sqrt{2}}\\ =\sqrt{2+1-2\sqrt{2}}\\ =\sqrt{{\left(\sqrt{2}-1\right)}^2}\\ =\sqrt{2}-1\end{array}$

So,

$N=\frac{\sqrt{2}\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\left(\sqrt{2}-1\right)$

Or

$\begin{array}{c}N=-\frac{\sqrt{2}\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\left(\sqrt{2}-1\right)\\ \Rightarrow N=\sqrt{2}-\sqrt{2}+1\end{array}$

Or

$\begin{array}{c}N=-\sqrt{2}-\sqrt{2}+1\\ \Rightarrow N=1\end{array}$

Or $N=-2\sqrt{2}+1$