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

 Apr 12, 2020
 #1
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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.

 Apr 12, 2020
 #2
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"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)

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

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

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

 Apr 14, 2020

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