# How do you solve x^4 + 2x + 2 = 0?

Aug 11, 2016

See explanation for an exploration of solving this quartic algebraically...

#### Explanation:

$f \left(x\right) = {x}^{4} + 2 x + 2$

Looks quite simple, does it not?

Solving quartics algebraically can get rather messy, even for an example as simple as this, but let's give it a try...

Since $f \left(x\right)$ is monic and has no cubic term, it can be factored as a product of monic quadratics with opposite middle coefficients:

${x}^{4} + 2 x + 2$

$= \left({x}^{2} - a x + b\right) \left({x}^{2} + a x + c\right)$

$= {x}^{4} + \left(b + c - {a}^{2}\right) {x}^{2} + \left(b - c\right) a x + b c$

Equating coefficients and rearranging a little we have:

$\left\{\begin{matrix}b + c = {a}^{2} \\ b - c = \frac{2}{a} \\ b c = 2\end{matrix}\right.$

Hence we find:

${\left({a}^{2}\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{4}{\left({a}^{2}\right)} + 8$

Multiplying both ends by $\left({a}^{2}\right)$ and rearranging slightly, we get a cubic in $\left({a}^{2}\right)$:

${\left({a}^{2}\right)}^{3} - 8 \left({a}^{2}\right) - 4 = 0$

This cubic has $3$ Real zeros, so if you try to solve it using Cardano's method you get solutions involving irreducible cube roots of Complex numbers (casus irreducibilis).

I will address this cubic in a separate question (see https://socratic.org/s/awX2qsH7), but suffice it to say that we are interested in the positive root (which I will call ${r}_{3}$ here):

${r}_{3} = \frac{4}{3} \sqrt{6} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right)\right) \approx 3.05$

Let $a = \sqrt{{r}_{3}}$

Then:

$b = \frac{1}{2} \left({a}^{2} + \frac{2}{a}\right) = {r}_{3} / 2 + \frac{1}{\sqrt{{r}_{3}}}$

$c = \frac{1}{2} \left({a}^{2} - \frac{2}{a}\right) = {r}_{3} / 2 - \frac{1}{\sqrt{{r}_{3}}}$

That leaves us with two quadratics to solve:

${x}^{2} - \sqrt{{r}_{3}} x + \left({r}_{3} / 2 + \frac{1}{\sqrt{{r}_{3}}}\right) = 0$

${x}^{2} + \sqrt{{r}_{3}} x + \left({r}_{3} / 2 - \frac{1}{\sqrt{{r}_{3}}}\right) = 0$

Using the quadratic formula, we find roots:

${x}_{1 , 2} = \frac{\sqrt{{r}_{3}} \pm \sqrt{{r}_{3} - 4 \left({r}_{3} / 2 + \frac{1}{\sqrt{{r}_{3}}}\right)}}{2}$

$= \frac{\sqrt{{r}_{3}} \pm \sqrt{{r}_{3} - 2 {r}_{3} - \frac{4}{\sqrt{{r}_{3}}}}}{2}$

$= \frac{1}{2} \left(\sqrt{{r}_{3}} \pm \left(\sqrt{{r}_{3} + \frac{4}{\sqrt{{r}_{3}}}}\right) i\right)$

$\textcolor{w h i t e}{}$

${x}_{3 , 4} = \frac{- \sqrt{{r}_{3}} \pm \sqrt{{r}_{3} - 4 \left({r}_{3} / 2 - \frac{1}{\sqrt{{r}_{3}}}\right)}}{2}$

$= \frac{- \sqrt{{r}_{3}} \pm \sqrt{{r}_{3} - 2 {r}_{3} + \frac{4}{\sqrt{{r}_{3}}}}}{2}$

$= \frac{1}{2} \left(- \sqrt{{r}_{3}} \pm \left(\sqrt{{r}_{3} - \frac{4}{\sqrt{{r}_{3}}}}\right) i\right)$