How do you find all the real and complex roots of #(x^3 + 9)(x^2 - 4) = 0#?

1 Answer
mason m
Mar 2, 2016

#x=-root3(9),+-2,root 3 9/2+-(3root6 3)/2i#

Explanation:

We can split this up into two separate equations:

#x^3+9=0" "" and "" "x^2-4=0#

Let's first solve the second equation (it's simpler):

#x^2-4=0#

#x^2=4#

#x=+-2#

These are the first #2# of what should be #5# total solutions.

Onto the next equation, we may be tempted to solve like so:

#x^3+9=0#

#x^3=-9#

#x=root3(-9)#

However, this doesn't deal with the two remaining zeros we should have. We must try to find a way to find the two leftover complex roots in #x^3+9=0#.

We can factor #x^3+9# as a sum of cubes, where the terms being cubed are #x# and #3^(2/3)#.

#x^3+9=0#

#(x+3^(2/3))(x^2-3^(2/3)x+3^(4/3))=0#

Solving the linear term gives us #x=-3^(2/3)=-root3 9#. To find the remaining two roots, use the quadratic formula on the quadratic factor.

#x=(3^(2/3)+-sqrt(3^(4/3)-4(1)(3^(4/3))))/2#

#x=(3^(2/3)+-sqrt(-3(3^(4/3))))/2#

#x=(3^(2/3)+-sqrt(3^(7/3))i)/2#

#x=3^(2/3)/2+-3^(7/6)/2i#

This can also be written as

#x=root 3 9/2+-(3root6 3)/2i#

or

#xapprox1.0400+-1.8014i#

depending on preference.

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1449 views around the world
You can reuse this answer
Creative Commons License