Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
 A: The general way to evaluate these kinds of expressions is to try to factor the contents of the square root into a perfect square.
One can try to find the factorization by letting $31 + 8\sqrt{15} = (a + b\sqrt{15})^2 = (a^2 + 15b^2) + 2ab\sqrt{15}$. We then test some possible values of $a$ and $b$. In this case, we see that $a = 4$ and $b = 1$ works. So, we can deduce that :
$$\begin{align}(4 + \sqrt{15})^2 &= 16 + 15 + 8 \sqrt{15} \\
&= 31 + 8 \sqrt{15}\end{align}$$
Similarly, we can deduce that 
$$\begin{align}(4 - \sqrt{15})^2 &= 16 + 15 - 8 \sqrt{15} \\
&= 31 - 8 \sqrt{15}\end{align}$$
Hence,
$$\begin{align}\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}} &= \sqrt{(4 - \sqrt{15})^2} + \sqrt{(4 + \sqrt{15})^2} \\
&= 4 - \sqrt{15} + 4 + \sqrt{15} \\ 
&= 8\end{align}$$
Of course, you can alternatively go ahead to obtain the solution by squaring the original expression, but it'll be rather tedious if you are working with bigger numbers.
A: You can square an expression, you know. Here you get
$(\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}})^2 = (31-8\sqrt{15}) + (31+8\sqrt{15}) + 2\sqrt{31^2-64.15} = 64$
So the answer is $8$.
A: Set up an equation like:
$$
u = \sqrt{31 -8 \sqrt{15}} + \sqrt{31 + 8 \sqrt{15}}
$$
Square to get it down to:
$$
u^2 = 31 - 8 \sqrt{15} + 31 + 8 \sqrt{15} + 2 \sqrt{31^2 - 8^2 \cdot 15}
    = 62 + 2 \cdot 1 = 64
$$
Thus you have $u = \pm 8$. The original expression is clearly positive, so:
$$
\sqrt{31 -8 \sqrt{15}} + \sqrt{31 + 8 \sqrt{15}} = 8
$$
A: Here is another way of doing it.
If we set $a=31+8\sqrt {15}, b=31-8\sqrt {15}$ we have 
$a+b=62$ and 
$ab=31^2-15\cdot 64=31^2-30\cdot32=1$
We know that $ab$ will work out well, because we know that $(x+y)(x-y)=x^2-y^2$
Then we know that $a$ and $b$ are the roots of $x^2-62x+1=0$
We want the square roots $\alpha$ and $\beta$ of $a$ and $b$ so if we set $y^2=x$ we will create an equation which has these as roots - $y^4-62y^2+1=0$
We know that $ab=1$ because we computed that. When we take the square root we get that $\alpha\beta=1$.
We therefore look for a factorisation $(y^2+py+1)(y^2-py+1)=y^4-62y^2+1$ which gives us $2-p^2=-62$, or $p^2=64$.
We know that one of these factors will have roots $\alpha, \beta$ and the other will have roots $-\alpha, -\beta$ so the final step is to work out which of these gives us the answer.
I have glossed over a couple of things because this is homework - but the double square root coming from a quartic without any terms in $y$ or $y^3$ is closely related to the underlying field theory, and is worth noticing.
