Question

How do you solve using the quadratic formula: `x^2+x+1=0`?

Answer 1

See explanation...

Explanation:

`x^2+x+1` is of the form `ax^2+bx+c` with `a = b = c = 1`

This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-(4xx1xx1)))/(2xx1)`

`=(-1+-sqrt(1-4))/2 = (-1+-sqrt(-3))/2 = (-1+-sqrt(3)i)/2`

`= -1/2+-sqrt(3)/2i`

The number `omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+sin((2pi)/3)i`

is called the primitive Complex cube root of unity and is used extensively when solving cubic equations.

`omega^2 = bar(omega) = -1/2-sqrt(3)/2i`

`omega^3 = 1`