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

My question here is that i go in an endless loop of substitution while using Vieta's formula. I want to keep using the Vieta formula so i can better understand it.

ive currently gotten this:

r + s = -4/3

rs = 4

but now im lost and dont know what to do.

im not asking for the answer, i would like a hint as to how to solve this by myself

 Dec 8, 2020
 #1
avatar+114221 
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3x^2  + 4x  + 12   =  0

 

You're  on  the right track....we  just need to use a little Algebra

 

r + s  = -4/3      square both sides

 

r^2  + 2rs  +  s^2 =  16/9      (1)

 

And

 

rs = 4  .....   so.....2rs  = 8      (2)

 

Sub (2)  into (1)    and you will find what you need

 

cool cool cool

 Dec 8, 2020
 #2
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tysm! i got it :>

Guest Dec 8, 2020

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