#3**+1 **

\(3x^2+4x+12=0\) 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

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

.Guest Apr 25, 2020

#4

#5**0 **

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