?siht evlos i od woh

To answer your other post:  the sum of the roots of a quadratic equation is equal to the negative of the linear term.

This is how he got his answer.

 

If you don't know this, or don't remember this, solve the function by using the quadratic formula (it's messy but you can do this!). Then square each term and add them together.

 

For me, I don't remember all these short-cuts, so I would do it the long way.

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

 

Notice \(r^2+s^2=(r+s)^2-2rs\) (1)

By Vieta's formulas:

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

\(rs=\frac{12}{3}=4\)

 

Now substitute into (1)

\((-\frac{4}{3})^2-2(4)\)

=\(\frac{16}{9}-8\)

\(=-\frac{56}{9}=-6.22222\)..