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r and s are the roots of 3x^2 + 4x - 12 = 0. Find r^2+s^2+2rs

 Jan 21, 2022
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Divide \(3x^2+4x-12=0\) by 3:  \(x^2+\frac{4}{3}x-4=0\)   ...(1)

 

Quadratic with roots r and s:  \((x-r)(x-s)=0\)  or  \(x^2-(r+s)x+rs=0\)    ...(2)

 

Compare (1) and (2)  to see that \(r+s=-\frac{4}{3}\)    ...(3)

 

Note that \(r^2 +s^2 + 2rs = (r+s)^2 \)

 

Now it should be easy to answer the question.

 Jan 21, 2022

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