# $3x^3 = 24$ quadratic equation

Completing the square I know by factoring

$$x^3 - 8 = 0\\ x-2 = 0$$

that one of the solutions is 2.

but the other solutions is $1 ± i \sqrt 3$.

Can someone explain to me how to get that?

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Is the only way to get it by using the quadratic formula? – CrewdNBasic Jun 5 '12 at 21:57
Divide out $x-2$ from $x^3 - 8$, then use the quadratic formula to solve the remaining quadratic. – Eugene Jun 5 '12 at 22:02
$x^3-8$ factors as $x^3-8=(x-2)(x^2+2x+4)$. The solutions to the equation $x^3-8=0$ are $x=2$, and the roots of $x^2+2x+4$. The easiest way to find the roots of the quadratic is to use the quadratic formula (which I think gives $-1\pm\sqrt 3 i$ as the roots). – David Mitra Jun 5 '12 at 22:02
One can sneak around the Quadratic Formula by completing the square, which amounts to proving the quadratic formula for this special quadratic. There are also more advanced ways of viewing the matter. – André Nicolas Jun 5 '12 at 22:06

May be you know this already but it's not clear to me from your post that you have the progression from the original problem statement to the final two factors exactly correct.

$3x^3 = 24 \\ \Rightarrow 3x^3 - 24 = 0 \\ \Rightarrow 3\left(x^3 - 8\right) = 0 \\ \Rightarrow x^3 - 8 = 0 \\ \Rightarrow \left(x-2\right)\left(x^2+2x+4\right) = 0 \\ \Rightarrow x-2 = 0, \ \ x^2+2x+4 = 0$

The first factor results in $x = 2$. The second factor must be solved using the quadratic formula or by completing the square since it is not factorable (Quadratic Formula on Wikipedia).

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Thanks! I got to the answer another way but this makes it a lot more clear! – CrewdNBasic Jun 5 '12 at 22:15

If we really want to avoid the Quadratic Formula, we can do it, by using more advanced ideas. For our problem, they are definitely overkill, but they are of great theoretical and practical importance.

Let $x=2w$. Our equation reduces to the equation $w^3=1$. By De Moivre's Formula, the (complex) solutions of the equation $w^n=1$, where $n$ is a positive integer, are $$w=\cos\left(\frac{2\pi k}{n}\right) +i\sin\left(\frac{2\pi k}{n}\right),$$ where $k$ ranges over the integers from $0$ to $n-1$.

For the case $n=3$, this says that the solutions are $w=\cos(0)+i\sin(0)$, $\cos(2\pi/3)+i\sin(2\pi/3)$, and $\cos(4\pi/3)+i\sin(4\pi/3)$.

The first solution is $w=0$. For the second solution, we need the sine and cosine of $2\pi/3$. These can be obtained from basic information about triangles. The same is true of the third solution.

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Hints

$$3x^{3}=24\Leftrightarrow x^{3}-8=0$$

$$\frac{x^{3}-8}{x-2}=x^{2}+2x+4$$

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Added: The second equation means that

$$x^{3}-8=\left( x-2\right) \left( x^{2}+2x+4\right) .$$

Thus we have

$$x^{3}-8=0\Leftrightarrow \left( x-2\right) \left( x^{2}+2x+4\right) =0,$$

whose solutions are the values of $x$ that make the factor $x-2$ or the factor $x^{2}+2x+4$ equal to $0$. The equation $x-2=0$ has the solution $x=2$. The solutions of the equation $x^{2}+2x+4=0$ may be found by the quadratic formula or as follows:

$$\begin{eqnarray*} x^{2}+2x+4 &=&0\Leftrightarrow \left( x+1\right) ^{2}+3=0\Leftrightarrow \left( x+1\right) ^{2}=-3 \\ &\Leftrightarrow &x+1=\pm \sqrt{-3}=\pm \sqrt{3}i \\ &\Leftrightarrow &x=-1\pm \sqrt{3}i. \end{eqnarray*}$$

So $$x^{3}-8=0\Leftrightarrow x=2\text{ or }x=-1\pm \sqrt{3}i.$$

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As David Mitra commented, one may factor using polynomial division:

$\frac {x^3 - 8} {x-2} = (x^2+2x+4)$

To solve this, we can quickly complete the square as follows:

Notice that we are setting $x^2 + 2x+4 = 0$

$(x+1)^2 + 3 = 0$

$(x+1)^2 = -3$

$x+1 = \pm i \sqrt{3}$

$x = -1 \pm i \sqrt{3}$ and we have our answer. I find this slightly faster than using the quadratic formula.

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does the x-2 just come from x^3-8 factored out? – CrewdNBasic Jun 5 '12 at 23:13
yes, it does. You're just factoring $x^3-8$ – Andrew Salmon Jun 6 '12 at 0:01

After changing variables $\rm\:x = 2\:\!z\:$ and cancelling $8$ we get $\rm\:0 = z^3 -1 = (z\!-\!1)\:\!(z^2\! +\! z\! +\! 1).\:$

Hence, by the quadratic formula, the nontrivial cube roots of $\:1\:$ are $\rm\:z = (-1\pm\sqrt{-3})/2.$

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