+0

# help!

0
112
2

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

Jun 5, 2020

#1
0

3x2 + 4x + 12  =  0

Use the quadratic formula to find the roots.

r  =  [ -4 + sqrt( 42 - 4·3·12 ) / ( 2·3 )  =  [ -4 + sqrt( -128 ) / ( 6 )   =  [ -4 + 8 · i · sqrt(2) ] / 6

s  =  [ -4 - sqrt( 42 - 4·3·12 ) / ( 2·3 )  =  [ -4 - sqrt( -128 ) / ( 6 )   =  [ -4 - 8 · i · sqrt(2) ] / 6

r2  =  ( [ -4 + 8 · i · sqrt(2) ] / 6 )2  =   [ -112 - 64·i·sqrt(2) ] / 36   (which can be reduced)

s2  =  ( [ -4 - 8 · i · sqrt(2) ] / 6 )2  =   ...

r2 + s2  =   ...

Jun 5, 2020
#2
+1

You CAN use the quadratic formula, but that takes a lot of time. Instead, you can use Vieta's Formulas. Vieta's for quadratics states that for a quadratic $$ax^2 + bx + c$$, the sum of the roots is $$-\frac{b}{a}$$ and the product of the roots is $$\frac{c}{a}$$. In case you want to find out more, here's an AoPS wiki link: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas

Anyway, $$r^2 + s^2 = (r+s)^2 - 2rs$$, which means that we can use Vieta's. Thus, $$r^2+s^2 = \left( -\frac{4}{3} \right)^2 - 2\cdot \frac{12}{3} = \frac{16}{9} - 8 = \boxed{-\frac{56}{9}}$$. Remember, when the problem is about some expression with the roots, if you can't factor the polynomial, try Vieta's.

Jun 5, 2020