+0

# need help

0
85
1

r and s are the roots of 3x^2 + 4x - 12 = 0. Find r^2+s^2+2rs

Jan 21, 2022

#1
+32957
+1

Divide $$3x^2+4x-12=0$$ by 3:  $$x^2+\frac{4}{3}x-4=0$$   ...(1)

Quadratic with roots r and s:  $$(x-r)(x-s)=0$$  or  $$x^2-(r+s)x+rs=0$$    ...(2)

Compare (1) and (2)  to see that $$r+s=-\frac{4}{3}$$    ...(3)

Note that $$r^2 +s^2 + 2rs = (r+s)^2$$

Now it should be easy to answer the question.

Jan 21, 2022