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How many complex roots does the equation below have? x6 + x3 + 1 = 0
Solution:
Given, the equation is x6 + x3 + 1 = 0
The equation can be written as (x3)2 + x3 + 1 = 0
Let us assume, x3 = t
So, the equation becomes
t2 + t + 1 = 0
Now, find the root using the quadratic formula
t = [-b ± √(b2 - 4ac)] / 2a
Here, a = 1, b = 1 and c = 1
t = [-(1) ± √((1)2 - 4(1)(1))] / 2(1)
t = [-1 ± √(1 - 4)] / 2
t = [-1 ± √-3] / 2
t = (-1 ± 3i) / 2
We know that, (-1 ± 3i) / 2 is a complex number.
⍵ = (-1 ± 3i) / 2
Now, x3 = ⍵
Taking cube root,
x = (⍵)1/3
Cube root of a complex number is complex.
Therefore, the number of complex roots is 6 as complex roots always occur in conjugate pairs.
How many complex roots does the equation below have? x6 + x3 + 1 = 0
Summary:
The number of complex roots of the equation x6 + x3 + 1 = 0 is 6. The number of complex roots is 6 as complex roots always occur in conjugate pairs.
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