Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
 A: $$
x = \sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}}\\
x^2 = \frac54 + \sqrt{\frac32} + 2\sqrt{\left(\frac54\right)^2 - \left(\sqrt{\frac32}\right)^2} + \frac54 - \sqrt{\frac32} = \frac52 + 2\sqrt{\frac{25}{16} - \frac32} = \frac52 + 2\cdot\frac14 = 3\\
x = \sqrt3
$$
A: $$\dfrac54\pm\sqrt{\dfrac32}=\dfrac{5\pm2\sqrt6}4=\left(\dfrac{\sqrt3\pm\sqrt2}2\right)^2$$
A: Why does squaring work here? Well it obviously deals with part of the square root. But there are two other things going on.
The first is that $$\left(\sqrt{a+\sqrt b}\right)^2+\left(\sqrt{a-\sqrt b}\right)^2=a+\sqrt b+a-\sqrt b=2a$$
And this means that the inner square roots disappear in part of the square.
The second is that $(a+\sqrt b)(a-\sqrt b)=a^2-b=c^2$. Here the cross term might have left a square root, but $c^2$ has a simple form.
Quite often in questions like this conjugates (expressions with the sign of the square root changed) come into play - the sum and product of two conjugate expressions both tend to be simple. So one technique for such questions is to look for a means to exploit that simplicity.

Note that the idea of conjugates can be generalised considerably, and is one of the ideas behind Galois Theory.
A: The general idea is to move along:
$$\sqrt{a}+\sqrt{r^2\,a} = \sqrt{a}+r\,\sqrt{a} = (1+r)\,\sqrt{a} = \sqrt{(1+r)^2\,a}$$
I am using the following systematic method as here, to first get the quotient:
$$r^2 = \frac{5/4+\sqrt{3/2}}{5/4-\sqrt{3/2}} = 49+20\sqrt{6} = (5+2\sqrt{6})^2$$
Therefore we have:
$$\sqrt{5/4-\sqrt{3/2}}+\sqrt{5/4+\sqrt{3/2}}= $$
$$\sqrt{(1+5+2\sqrt{6})^2\,(5/4-\sqrt{3/2})} = $$
$$\sqrt{(60+24\sqrt{6})\,(5/4-\sqrt{3/2})} = \sqrt{3}$$
