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

 

thank you!

 Oct 12, 2020
 #1
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By Vieta's formulas, r + s = -4 and rs = 12.  Squaring the equation r + s = -4, we get r^2 + 2rs + s^2 = 16.  Then r^2 + s^2 = 16 - 2rs = 16 - (2rs) = 16 - 2(12) = -8.

 Oct 12, 2020
 #2
avatar+19 
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Hey, thanks for the answer, It appears it isnt correct, and I dont know what to do. I've used Vietas formula on this so many times and ive came up with -4, and apparently -8 isnt correct, thank you for the answer, can anybody else help?

Age26  Oct 12, 2020
 #3
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As follows

 Oct 12, 2020

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