# How do you find all solutions to x^5+243=0?

##### 2 Answers
Sep 18, 2016

$x \in \left\{- 3 {e}^{i \frac{2 \pi n}{5}} | n \in \left\{0 , 1 , 2 , 3 , 4\right\}\right\}$

#### Explanation:

We will make use of the fact that ${e}^{i \theta} = {e}^{i \left(\theta + 2 \pi k\right)}$ where $k$ is an integer (this is clear from Euler's formula e^(itheta) = cos(theta)+isin(theta)), along with that ${e}^{0} = 1$.

${x}^{5} + 243 = 0$

$\implies {x}^{5} = - 243 = - 243 {e}^{i \left(0 + 2 \pi k\right)} , k \in \mathbb{Z}$

$\implies x = {\left(- 243 {e}^{i 2 \pi k}\right)}^{\frac{1}{5}} = - 3 {e}^{i \frac{2 \pi k}{5}}$

Due to the periodic nature of ${e}^{i \theta}$, we have that for any $k \in \mathbb{Z}$, ${e}^{i \frac{2 \pi k}{5}} = {e}^{i \frac{2 \pi n}{5}}$ for some $n \in \left\{0 , 1 , 2 , 3 , 4\right\}$. Thus, we can find our five possible values for $x$ by substituting in each possible value for $n$.

$x \in \left\{- 3 {e}^{i \frac{2 \pi n}{5}} | n \in \left\{0 , 1 , 2 , 3 , 4\right\}\right\}$

Note that as ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, and $\sin \left(\frac{2 \pi n}{5}\right) = 0$ only when $n = 0$, the only real solution to the equation is $- 3 {e}^{i \frac{2 \pi \cdot 0}{5}} = - 3 {e}^{0} = - 3$

Sep 18, 2016

Solutions in $a + b i$ form:

$x = - 3$

$x = \frac{3 \left(1 + \sqrt{5}\right)}{4} \pm \frac{3 \sqrt{10 - 2 \sqrt{5}}}{4} i$

$x = \frac{3 \left(1 - \sqrt{5}\right)}{4} \pm \frac{3 \sqrt{10 + 2 \sqrt{5}}}{4} i$

#### Explanation:

Here's an alternative approach to find the solutions in $a + b i$ form...

color(white)
Fifth roots of $1$

First let us find the Complex fifth roots of $1$ in $a + b i$ form:

${x}^{5} - 1 = \left(x - 1\right) \left({x}^{4} + {x}^{3} + {x}^{2} + x + 1\right)$

$\frac{1}{{x}^{2}} \left({x}^{4} + {x}^{3} + {x}^{2} + x + 1\right) = {x}^{2} + x + 1 + \frac{1}{x} + \frac{1}{x} ^ 2$

$\textcolor{w h i t e}{\frac{1}{{x}^{2}} \left({x}^{4} + {x}^{3} + {x}^{2} + x + 1\right)} = {\left(x + \frac{1}{x}\right)}^{2} + \left(x + \frac{1}{x}\right) - 1$

Using the quadratic formula we find zeros:

$x + \frac{1}{x} = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

Hence:

${x}^{2} + \left(\frac{1}{2} \pm \frac{\sqrt{5}}{2}\right) x + 1 = 0$

$\textcolor{w h i t e}{}$
Case $\boldsymbol{{x}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) x + 1 = 0}$

Using the quadratic formula:

$x = \frac{- \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) \pm \sqrt{{\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{2} - 4}}{2}$

$\textcolor{w h i t e}{x} = \frac{- \left(1 + \sqrt{5}\right) \pm \sqrt{{\left(1 + \sqrt{5}\right)}^{2} - 16}}{4}$

$\textcolor{w h i t e}{x} = \frac{- \left(1 + \sqrt{5}\right) \pm \sqrt{2 \sqrt{5} - 10}}{4}$

$\textcolor{w h i t e}{x} = - \frac{1 + \sqrt{5}}{4} \pm \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i$

$\textcolor{w h i t e}{}$
Case $\boldsymbol{{x}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) x + 1 = 0}$

Similarly (or simply reversing the signs of the coefficients of $\sqrt{5}$)

$x = - \frac{1 - \sqrt{5}}{4} \pm \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i$

$\textcolor{w h i t e}{}$
Solve $\boldsymbol{{x}^{5} + 243} = 0$

Since $243 = {3}^{5}$, the solutions of this are $x = - 3 \alpha$ for any $\alpha$ which is a fifth root of $1$.

So the solutions are:

$x = - 3$

$x = \frac{3 \left(1 + \sqrt{5}\right)}{4} \pm \frac{3 \sqrt{10 - 2 \sqrt{5}}}{4} i$

$x = \frac{3 \left(1 - \sqrt{5}\right)}{4} \pm \frac{3 \sqrt{10 + 2 \sqrt{5}}}{4} i$

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

Since we were able to construct the $5$th roots of $1$ using basic arithmetic operations and square roots, it means that a regular pentagon can be constructed using a straight edge and compass.