+0

# Algebra

0
117
2

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

$$\phantom{cube is 343}$$

May 22, 2022

#1
+9461
0

Suppose that the non-real root is $$\omega$$. Obviously, since it is non-real, it cannot be equal to 1.

Since its cube is 1, $$\omega^3 = 1$$, which means $$\omega^3 - 1 = 0$$. If we factorize the left-hand side of the equation, we have $$(\omega - 1)(\omega^2 + \omega + 1) = 0$$. Since $$\omega$$ is not 1, we can cross out $$(\omega - 1)$$, so we're left with $$\omega^2 + \omega + 1 = 0$$.

With the derivation above, it turns out that if $$\omega$$ is a non-real root whose cube is 1, then $$x = \omega$$ must be a root of the equation $$x^2 + x + 1 = 0$$. Therefore, a = b = 1 by comparing coefficients.

May 23, 2022
#2
+9461
0

I used this symbol for the non-real root for a reason. It is called the "cube roots of unity". Generally, any root of unity is commonly denoted by $$\omega$$

WIki page: https://en.wikipedia.org/wiki/Root_of_unity

MaxWong  May 23, 2022