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r and s are the roots of 3x^2 + 4x - 12 = 0. Find r^2+s^2 Please help me!!!

 Jun 27, 2021
 #1
avatar+121000 
+2

By Vieta

 

The  sum of the roots  =     -4/3    = r +  s

Square  both sides

16/9  =  r^2  +  2rs  + s^2             (1)

 

The product of the  roots =   rs =  -12/3   =   -4

So  2rs    =   -8           (2)

 

Sub  (2)  into (1)   and we  have

r^2  - 8  +  s^2  =   16/9

r^2  +  s^2   =   8 + 16/9

r^2  + s^2   =  72/9  + 16/9

r^2  + s^2   =  88  / 9

 

 

cool cool cool

 Jun 27, 2021
 #2
avatar+732 
+1

$r^2+s^2 = (r+s)^2 - 2rs$

$r+s = - \frac{4}{3}$

$rs = -4$

\(\begin{align*} r^2 + s^2 &= \left(- \frac{4}{3} \right)^2 - 2 (-4) \\ &= \frac{16}{9} + 8 \\ &= \boxed{\frac{88}{9}} \end{align*}\)

 Jun 27, 2021

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