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

3 Answers
Jan 23, 2016

Answer:

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:

#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:

#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...