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! :)
 A: Hint:
$$4\sqrt3+8 = (\sqrt2+\sqrt6)^2$$
A: 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 $x_4$
A: $$x^4-8\sqrt{3}x^2-16=0\Longleftrightarrow$$

Substitute $y=x^2$:

$$y^2-8\sqrt{3}y-16=0\Longleftrightarrow$$
$$y^2-8\sqrt{3}y=16\Longleftrightarrow$$
$$y^2-8\sqrt{3}y+16\cdot 3^{2\cdot\frac{1}{2}}=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$
$$\left(y-4\sqrt{3}\right)^2=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3^{2\cdot\frac{1}{2}}}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm\sqrt{64}\Longleftrightarrow$$
$$y-4\sqrt{3}=\pm 8\Longleftrightarrow$$
$$y=\pm 8+4\sqrt{3}\Longleftrightarrow$$
$$x^2=\pm 8+4\sqrt{3}\Longleftrightarrow$$
$$x=\pm\sqrt{\pm 8+4\sqrt{3}}$$
So the solutions are:
$$x_1=\sqrt{8+4\sqrt{3}}=\sqrt{2}+\sqrt{6}\approx 3.86370$$
$$x_2=-\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{2}-\sqrt{6}\right)\approx -1.0325i$$
$$x_3=\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{6}-\sqrt{2}\right)\approx 1.0325i$$
$$x_4=-\sqrt{8+4\sqrt{3}}=-\left(\sqrt{2}+\sqrt{6}\right)\approx -3.86370$$
A: You got to $x^2 = 4\sqrt 3 \pm 8$
Since $4\sqrt 3 - 8 < 0$, then $x^2 = 4\sqrt 3 + 8$. What can you do if you cant see that $(\sqrt 2 + \sqrt 6)^2 = 4 \sqrt 3 + 8$?
Let $$\text{$x^2 = 8 + 4\sqrt 3 \ $ and $ \ y^2 = 8 - 4\sqrt 3$}$$
We can assume that $x > y > 0$.
Then $x^2y^2 = 8^2 - (4\sqrt 3)^2 = 16$. So $xy = 4$.
$(x+y)^2 = (x^2+y^2) + 2xy = 16 + 8 = 24$. So $x+y = 2\sqrt 6$.
$(x-y)^2 = (x^2+y^2) - 2xy = 16 - 8 = 8$. So $x+y = 2\sqrt 2$.
So
\begin{align}
   (x+y)+(x-y) &= 2\sqrt 6 + 2 \sqrt 2 \\
   2x &= 2\sqrt 6 + 2 \sqrt 2 \\
   x &= \sqrt 6 + \sqrt 2
\end{align}
Since $x$ can also be negative,  then $x = \pm(\sqrt 6 + \sqrt 2)$
We compute 
$(x + (\sqrt 6 + \sqrt 2))(x - (\sqrt 6 + \sqrt 2)) = x^2 - 8 + 4\sqrt 3$
We find that
$x^4- 8\sqrt{3}x^2 - 16 = (x^2 - 4\sqrt 3 - 8)(x^2 - 4\sqrt 3 + 8)$
It only seems proper to check that 
$(\sqrt 6 - \sqrt 2)^2 = 8 - 4\sqrt 3$
So the roots are 
$$\{\pm(\sqrt 6 + \sqrt 2), \pm i(\sqrt 6 - \sqrt 2)   \}$$
