Let r and s be the roots of $3x2+4x−12=0.
Find r2+s2
Sum of the roots = r + s = -4/3 square both sides
r^2 + 2rs + s^2 = 16/9
Product of the roots = rs = -12/3 = 4
2rs = 8
So
r^2 + 8 + s^2 = 16/9
r^2 + s^2 = 16/9 - 8
r^2 + s^2 = -56/9
