Question

Let $r$ and $s$ be the roots of $3x^2 + 4x - 12 = 0.$ Find $r^2 + s^2.$

Answer 1

We can factorize the quadratic as

$3x^2 + 4x - 12 = 3 (x - r) (x - s)$

and expand the right side to get

$3x^2 + 4x - 12 = 3x^2 - 3(r + s) x + 3rs$

$\implies r + s = -\dfrac43 \text{ and } rs = -4$

Then we find

$(r + s)^2 = r^2 + 2rs + s^2 \implies r^2 + s^2 = \left(-\dfrac43\right)^2 - 2(-4) = \boxed{\dfrac{88}9}$