How to simplify this surd: $\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$ $$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$
We have to find the value of $x$. 
Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.
I think the answer lies in simplification, but can't do it. Also I have tried taking $\sqrt{2}$ common, but it doesn't help.

try converting expr inside surd into a square.
$$x = \sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$$
$$ = \sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}} $$
$$ = \sqrt{\frac{1+3+2\sqrt{3}}{2^2}}+\sqrt{\frac{1+3-2\sqrt{3}}{2^2 }} $$
$$ = \sqrt{\left(\frac{1+\sqrt{3}}{2}\right)^2}+\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^2} $$
$$ = \frac{|1+\sqrt{3}|}{2} + \frac{|1-\sqrt{3}|}{2}$$ 
$$ = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}$$ 
$$ = \sqrt{3}$$
Edit
My earlier answer was incorrect since i foolishly took square root and forgot about modulus. I have ammended it. It is important to note that: 
$$\sqrt{x^2} = |x|$$
