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

3 Answers
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 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.

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

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...