Question

if r and s are solutions of x^2 + 6x - 2 = 0, what is the value of r^3 + s^3.

Answer 1

use the quadratic formula to find the roots of the quadratic

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

a = 1, b = 6 and c = -2

the two values you get for x... will be r and s... then cube each and add the answers