<s>[INST] 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! :)
 [/INST] $$x^4-8\sqrt{3}x^2-16=0\Longleftrightarrow$$

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:
$$x_1=\sqrt{8+4\sqrt{3}}=\sqrt{2}+\sqrt{6}\approx 3.86370$$
$$x_2=-\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{2}-\sqrt{6}\right)\approx -1.0325i$$
$$x_3=\sqrt{-8+4\sqrt{3}}=i\left(\sqrt{6}-\sqrt{2}\right)\approx 1.0325i$$
$$x_4=-\sqrt{8+4\sqrt{3}}=-\left(\sqrt{2}+\sqrt{6}\right)\approx -3.86370$$
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