A hard square root question

Question

This is my first question on StackExchange. So my question is:

If $$x = \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{\sqrt{\sqrt5 + 1}} + \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{2\sqrt{\sqrt5 + 1}} - \sqrt{3-2\sqrt2}$$ What is the value of $x^2$?

If someone can also tell me how to input mathematical functions on the StackExchange it would be appreciated.

Answer 1

Letting $$\frac{\sqrt{\sqrt 5+2}+\sqrt{\sqrt 5-2}}{\sqrt{\sqrt 5+1}}=\alpha,$$ we have $$\alpha^2=\frac{(\sqrt 5+2)+(\sqrt 5-2)+2\sqrt{(\sqrt 5+2)(\sqrt 5-2)}}{\sqrt 5+1}=\frac{2(\sqrt 5+1)}{\sqrt 5+1}=2\Rightarrow \alpha=\sqrt 2.$$ Hence, we have $$x=\alpha+\frac{\alpha}{2}-\sqrt{(\sqrt 2-1)^2}=\frac{3}{2}\alpha-(\sqrt 2-1)=\frac{\sqrt 2}{2}+1\Rightarrow x^2=\frac{3}{2}+\sqrt 2.$$

Answer 2

I'd say the fastest way to do this is type it into a calculator (or some mathematical tool like matlab or R)

I might very likely have made a typo when entering your formula but my results are

• $x\approx3.288246$
• $x^2\approx10.81256$

please check it yourself!

Answer 3

Hint: $\sqrt{\sqrt5+2}\cdot\sqrt{\sqrt5-2}~=~?$, and $\sqrt{\sqrt5+1}\cdot\sqrt{\sqrt5-1}~=~?$ Also, if $(a-b)^2=a^2+b^2$

$-2ab=3-2\sqrt2$, then what are the values of a and b ?