Solving the 4th Degree Equation $x^4- 8\sqrt{3}x^2 - 16 = 0$ I'm learning radical simplification and our teacher gave us this equation to solve:
$$x^4-8\sqrt{3}x^2-16=0$$
She told us to consider $y=x^2$ to transform the equation into a quadratic equation, which we can solve. However, when I apply the quadratic formula to the equation:
$$y^2-8\sqrt{3}y-16=0$$
I get:
$$y={8\sqrt{3}\pm\sqrt{\sqrt{192}-64}\over 2}\equiv y=4\sqrt{3}\pm8$$
And after that I get stuck. Our teacher solved another equation in class and she transformed the result of the quadratic formula into the square of a binomial, so that you're able to square root it and get the value of $x$, however, I haven't been able to transform $y=4\sqrt{3}\pm8$ into a square of a binomial. Maybe there's another way around it, but I can't seem to find it.
Any help would be greatly appreciated, thanks in advance! :)

You have 
$x^4-8\sqrt3x^2-16=(x^2-4\sqrt3)^2-48-16=0\Rightarrow x^2=4\sqrt3\pm8\Rightarrow x=\pm\sqrt{4\sqrt3\pm8}$
Hence $\begin {cases}x_1=\sqrt{4\sqrt3 +8}\\x_2=\sqrt{4\sqrt 3-8}\\x_3=-x_1\\x_4=-x_2\end{cases}$
Make now, if you want to simplify the irrational of degree four,
$\sqrt{4\sqrt 3+8}=\sqrt a+\sqrt b$ which gives the system $a+b=8$ and $2\sqrt{ab}=4\sqrt 3$. It implies the equation $a^2-8a+12=(a-2)(a-6)=0$ thus $\sqrt{4\sqrt 3+8}=\sqrt 2+\sqrt 6$ and you can finish taking care on the non real roots in $x_2$ and $x_4$
