Solving the 4th Degree Equation $x^4- 8\sqrt{3}x^2 - 16 = 0$

Question

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! :)

Answer 1

Hint:

$$4\sqrt3+8 = (\sqrt2+\sqrt6)^2$$

Answer 2

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 $x_4$

Answer 3

$$x^4-8\sqrt{3}x^2-16=0\Longleftrightarrow$$

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Substitute $y=x^2$:

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$$y^2-8\sqrt{3}y-16=0\Longleftrightarrow$$ $$y^2-8\sqrt{3}y=16\Longleftrightarrow$$ $$y^2-8\sqrt{3}y+16\cdot 3^{2\cdot\frac{1}{2}}=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$ $$\left(y-4\sqrt{3}\right)^2=16+16\cdot 3^{2\cdot\frac{1}{2}}\Longleftrightarrow$$ $$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3^{2\cdot\frac{1}{2}}}\Longleftrightarrow$$ $$y-4\sqrt{3}=\pm\sqrt{16+16\cdot 3}\Longleftrightarrow$$ $$y-4\sqrt{3}=\pm\sqrt{64}\Longleftrightarrow$$ $$y-4\sqrt{3}=\pm 8\Longleftrightarrow$$ $$y=\pm 8+4\sqrt{3}\Longleftrightarrow$$ $$x^2=\pm 8+4\sqrt{3}\Longleftrightarrow$$ $$x=\pm\sqrt{\pm 8+4\sqrt{3}}$$

So the solutions are:

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