# How do you find all the real and complex roots of x^5 - 32 = 0?

Jan 23, 2016

You must mean the solutions to the equation. In the solution below I will show you how to solve this equation

#### Explanation:

${x}^{5}$ - 32 = 0

${x}^{5}$ = 32

Now, since ${x}^{2}$ is opposite to √x, and ${x}^{3}$ is opposite to $\sqrt[3]{x}$, ${x}^{5}$ must be opposite to $\sqrt[5]{x}$.

x = $\sqrt[5]{32}$

x = 2

Since an odd exponent with a negative base always gives a negative number (example $- {2}^{5}$ = -32), x = 2 is the only possible solution to this equation.

Exercises:

1. Solve for x in the following equations.

a) ${x}^{3}$ - 27 = 0

b) ${x}^{4}$ - 16 = 0

Hopefully you understand now, and best of luck in your continuation.

Feb 6, 2016

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

#### Explanation:

It is evident that $x = 2$ is a root since $32 = {2}^{5}$. The real trick here is finding the remaining $4$ complex roots.

Since $x = 2$ is a root, $\left(x - 2\right)$ is a factor of the quintic. The remaining roots can be found through solving the remaining quartic:

$\frac{{x}^{5} - 32}{x - 2} = {x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16$

We want to solve for $x$:

${x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16 = 0$

I will admit that finding the complex roots is difficult and unintuitive (I resorted to using Google to find a way to factor this):

Notice that ${\left({x}^{2} + x + 4\right)}^{2} = {x}^{4} + 2 {x}^{3} + 9 {x}^{2} + 8 x + 16$, which is almost the quartic we had remaining.

Thus, we can say that

${x}^{4} + 2 {x}^{3} + 9 {x}^{2} + 8 x + 16 - 5 {x}^{2} = 0$

Which simplifies to be

${\left({x}^{2} + x + 4\right)}^{2} - 5 {x}^{2} = 0$

This can be factored as a difference of squares.

$\left({x}^{2} + x + 4 + x \sqrt{5}\right) \left({x}^{2} + x + 4 - x \sqrt{5}\right) = 0$

These quadratic equations can be solved individually:

The first:

${x}^{2} + \left(1 + \sqrt{5}\right) x + 4 = 0$

$x = \frac{- 1 - \sqrt{5} \pm \sqrt{{\left(1 + \sqrt{5}\right)}^{2} - 16}}{2}$

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

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

The second:

${x}^{2} + \left(1 - \sqrt{5}\right) x + 4 = 0$

$x = \frac{\sqrt{5} - 1 \pm \sqrt{{\left(1 - \sqrt{5}\right)}^{2} - 16}}{2}$

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

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

Feb 7, 2016

$x = 2 {e}^{\frac{2 k \pi i}{5}} , k = 0 , 1 , 2 , 3 , 4$

#### Explanation:

${x}^{5} - 32 = 0$

${x}^{5} = 32$

$= \sqrt[5]{32}$

The real solution is obviously 2.

Let's extend the equation to polar complexes:

${x}^{5} = 32 = 32 {e}^{2 k \pi i}$, with $k$ any integer, but pratically, from 0 to 4.

$x = \sqrt[5]{32 {e}^{2 k \pi i}}$

$x = 2 {e}^{\frac{2 k \pi i}{5}} , k = 0 , 1 , 2 , 3 , 4$

MORE EXPLANATION OR ANOTHER WAY OF LOOKING AT IT:

An easy way to see this to imagine that the solution divided up around the circle of radius 2 in complex plane. Now divide the circle by 5 so the solutions will be: ${x}_{1} = 2 \angle \theta$

1) Real ${x}_{1} = 2 \angle \left(\theta = 0\right)$
2) Complex ${x}_{2} = 2 \angle \left(\theta = \frac{2 \pi}{5}\right)$
Find the real and imaginary part: $2 \cos \theta + 2 i \sin \theta$
R = 2cos((2pi)/5); I = sin((2pi)/5)
3) Complex ${x}_{3} = 2 \angle \left(\theta = \frac{4 \pi}{5}\right)$ this is $2 \left(\frac{2 \pi}{5}\right)$
Find the real and imaginary part: $2 \cos \theta + 2 i \sin \theta$
4) Complex ${x}_{4} = 2 \angle \left(\theta = \frac{6 \pi}{5}\right)$ this is $3 \left(\frac{2 \pi}{5}\right)$
Find the real and imaginary part: $2 \cos \theta + 2 i \sin \theta$
4) Complex ${x}_{5} = 2 \angle \left(\theta = \frac{8 \pi}{5}\right)$ this is $4 \left(\frac{2 \pi}{5}\right)$
Find the real and imaginary part: $2 \cos \theta + 2 i \sin \theta$

Hope this help...