# 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! :)

Hint:

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

• What about the solution where $y=4\sqrt{3}-8$? How can you get -8 from $\sqrt{2}$ and $\sqrt{6}$? – António Bezerra Nov 21 '15 at 13:00
• $4\sqrt{3}-8<0$ – Claude Leibovici Nov 21 '15 at 13:11
• If you like, $4\sqrt3-8 = i^2(\sqrt2-\sqrt6)^2$, which would help in finding complex roots. – Macavity Nov 21 '15 at 13:26
• Since we haven't learned about complex numbers I'm guessing we're supposed to ignore the $4\sqrt{3}-8$ solution. – António Bezerra Nov 21 '15 at 13:54
• Sure, the positive root of $y$ then will give you the possible values of $x$ which are real. – Macavity Nov 21 '15 at 14:02

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 $x_4$

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

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:

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

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 $$\{\pm(\sqrt 6 + \sqrt 2), \pm i(\sqrt 6 - \sqrt 2) \}$$