Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential. 
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards

In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+b\sqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1 
($a+b\sqrt 3)^4=28+16\sqrt 3$. So suppose ($a+b\sqrt 3)^2=c+d\sqrt 3$. Then $(c+d\sqrt 3)^2=28+16\sqrt 3$, i.e.  
$$(c^2+3d^2)+2cd\sqrt 3=28+16\sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+\dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=\pm4$, and so $d=\pm2$. So we have two possible solutions: $\pm(4+2\sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16\sqrt 3$). Pick the positive solution $4+2\sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2 
Now you need to solve $(a+b\sqrt 3)^2=4+2\sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
