+0  
 
0
112
2
avatar

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

 Jun 5, 2020
 #1
avatar+21955 
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
avatar+6 
+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

9 Online Users

avatar