Question

How many complex roots does the equation below have? X^6+x^3+1=0

Answer 1

If you find the discriminant it will tell you the number and types of roots. The discriminant is the value b^2 -4ac.
a = 1
b = 1
c = 1
1^2 - 4*1*1
1-4 = -3
Since this is a negative number there will be 2 complex roots.

Answer 2

Answer:

The number of complex roots is 6.

Step-by-step explanation:

The given equation is

$x^6+x^3+1=0$

it can be written as

$(x^3)^2+x^3+1=0$

Substitute $t=x^3$,

$(t)^2+t+1=0$

Using quadratic formula.

$t=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2}=\frac{-1\pm 3i}{2}$     $t=\frac{-b\pm \sqrt{b^2-4ac}}{2}$

We know that

$\omega =\frac{-1\pm 3i}{2}$

It is a complex number.

$x^3=\omega$

$x=(\omega)^{\frac{1}{3}}$

Cube root of a complex number is complex.

Therefore the number of complex roots is 6.