Question

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

Answer 1

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 `root(3)x`, `x^5` must be opposite to `root(5)x`.

x = `root(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.

Answer 2

`x=2,(-1-sqrt5+-isqrt(10-2sqrt5))/2,(sqrt5-1+-isqrt(10+2sqrt5))/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, `(x-2)` is a factor of the quintic. The remaining roots can be found through solving the remaining quartic:

`(x^5-32)/(x-2)=x^4+2x^3+4x^2+8x+16`

We want to solve for `x`:

`x^4+2x^3+4x^2+8x+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 `(x^2+x+4)^2=x^4+2x^3+9x^2+8x+16`, which is almost the quartic we had remaining.

Thus, we can say that

`x^4+2x^3+9x^2+8x+16-5x^2=0`

Which simplifies to be

`(x^2+x+4)^2-5x^2=0`

This can be factored as a difference of squares.

`(x^2+x+4+xsqrt5)(x^2+x+4-xsqrt5)=0`

These quadratic equations can be solved individually:

The first:

`x^2+(1+sqrt5)x+4=0`

`x=(-1-sqrt5+-sqrt((1+sqrt5)^2-16))/2`

`x=(-1-sqrt5+-sqrt(-10+2sqrt5))/2`

`x=(-1-sqrt5+-isqrt(10-2sqrt5))/2`

The second:

`x^2+(1-sqrt5)x+4=0`

`x=(sqrt5-1+-sqrt((1-sqrt5)^2-16))/2`

`x=(sqrt5-1+-sqrt(-10-2sqrt5))/2`

`x=(sqrt5-1+-isqrt(10+2sqrt5))/2`

Answer 3

`x=2e^((2kpii)/5), k=0,1,2,3,4`

Explanation:

`x^5-32=0`

`x^5=32`

`=root(5)(32)`

The real solution is obviously 2.

Let's extend the equation to polar complexes:

`x^5=32=32e^(2kpii)`, with `k` any integer, but pratically, from 0 to 4.

`x=root(5)(32e^(2kpii))`

`x=2e^((2kpii)/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/_ theta`

1) Real `x_1 = 2/_ (theta = 0)`
2) Complex `x_2 = 2/_ (theta = (2pi)/5)`
Find the real and imaginary part: `2costheta + 2isintheta`
`R = 2cos((2pi)/5); I = sin((2pi)/5)`
3) Complex `x_3 = 2/_ (theta = (4pi)/5)` this is `2((2pi)/5)`
Find the real and imaginary part: `2costheta + 2isintheta`
4) Complex `x_4 = 2/_ (theta = (6pi)/5)` this is `3((2pi)/5)`
Find the real and imaginary part: `2costheta + 2isintheta`
4) Complex `x_5 = 2/_ (theta = (8pi)/5)` this is `4((2pi)/5)`
Find the real and imaginary part: `2costheta + 2isintheta`

Hope this help...