Question

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

Answer 1

Answer:

-56/9

Step-by-step explanation:

By Vieta's formulas,

$r + s = -\frac{4}{3}$ and $rs = \frac{12}{3} = 4.$ Squaring the equation $r + s = -\frac{4}{3},$ we get

$r^2 + 2rs + s^2 = \frac{16}{9}.$ Therefore,

$r^2 + s^2 = \frac{16}{9} - 2rs = \frac{16}{9} - 2 \cdot 4 = -\frac{56}{9}}$