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

 Apr 25, 2020
 #1
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This problem was recently answered; please check below.

 Apr 25, 2020
 #2
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where?????

 Apr 25, 2020
 #3
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\(3x^2+4x+12=0\) Is a quadratic equation of the form \(ax^2+bx+c=0\) where r and s are roots 

By Vieta's formulas:

\(r+s=-\frac{b}{a}\) Thus \(r+s=-\frac{4}{3}\)

\(r*s=\frac{c}{a}\) Thus \(rs=4\)

Now notice that:

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

Substitute 

\((-\frac{4}{3})^2-2(4)=-\frac{56}{9}=-6.22\)

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 Apr 25, 2020
 #4
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it is wrong?????

 Apr 26, 2020
 #5
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Good, I hope it is, you are too lazy to even lay your question out properly. 

And too lazy to search old answers.

I would have preferred that you were not answered at all but if the answer is wrong that is the next best thing.

 

(No disrespect intended to answering guest, I know you are just trying to help)

Melody  Apr 26, 2020

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