Find the ordered pair (a,b) of real numbers for which x^2+ax+b has a non-real root whose cube is 1.

Guest Jun 9, 2022

#1**0 **

Let the root be \(z\).

We have the equation: \(z^3 = 1 \)

Subtracting 1 from both sides gives: \(z^3 - 1 = 0\)

Factoring out the left-hand side, we get: \((z-1)(z^2 + z + 1) = 0 \)

Oh, look... It's a quadratic that so happens to be in the form \(ax^2 + bx + c = 0 \)

This means that the quadratic we are looking for is \(z^2 + z + 1 = 0 \), so the ordered pair is \(\color{brown}\boxed{(1,1)}\)

BuilderBoi Jun 9, 2022