How do you find all the real and complex roots of (x^3 + 9)(x^2 - 4) = 0(x3+9)(x2−4)=0?
1 Answer
Explanation:
We can split this up into two separate equations:
x^3+9=0" "" and "" "x^2-4=0x3+9=0 and x2−4=0
Let's first solve the second equation (it's simpler):
x^2-4=0x2−4=0
x^2=4x2=4
x=+-2x=±2
These are the first
Onto the next equation, we may be tempted to solve like so:
x^3+9=0x3+9=0
x^3=-9x3=−9
x=root3(-9)x=3√−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
We can factor
x^3+9=0x3+9=0
(x+3^(2/3))(x^2-3^(2/3)x+3^(4/3))=0(x+323)(x2−323x+343)=0
Solving the linear term gives us
x=(3^(2/3)+-sqrt(3^(4/3)-4(1)(3^(4/3))))/2x=323±√343−4(1)(343)2
x=(3^(2/3)+-sqrt(-3(3^(4/3))))/2x=323±√−3(343)2
x=(3^(2/3)+-sqrt(3^(7/3))i)/2x=323±√373i2
x=3^(2/3)/2+-3^(7/6)/2ix=3232±3762i
This can also be written as
x=root 3 9/2+-(3root6 3)/2ix=3√92±36√32i
or
xapprox1.0400+-1.8014ix≈1.0400±1.8014i
depending on preference.
