r and s are the roots of 3x^2 + 4x - 12 = 0. Find r^2+s^2 Please help me!!!
+129850 By Vieta
The sum of the roots = -4/3 = r + s
Square both sides
16/9 = r^2 + 2rs + s^2 (1)
The product of the roots = rs = -12/3 = -4
So 2rs = -8 (2)
Sub (2) into (1) and we have
r^2 - 8 + s^2 = 16/9
r^2 + s^2 = 8 + 16/9
r^2 + s^2 = 72/9 + 16/9
r^2 + s^2 = 88 / 9
+876 $r^2+s^2 = (r+s)^2 - 2rs$
$r+s = - \frac{4}{3}$
$rs = -4$
\(\begin{align*} r^2 + s^2 &= \left(- \frac{4}{3} \right)^2 - 2 (-4) \\ &= \frac{16}{9} + 8 \\ &= \boxed{\frac{88}{9}} \end{align*}\)
