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Let r and s be the roots of 3x^2 + 4x + 12 = 2x^2 - 7x + 5.  Find r^2 + s^2.

 May 24, 2021
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Let r and s be the roots of \(3x^2 + 4x + 12 = 2x^2 - 7x + 5\).
Find \(r^2 + s^2\).

 

\(\begin{array}{|rcll|} \hline \mathbf{ 3x^2 + 4x + 12 } &=& \mathbf{ 2x^2 - 7x + 5 } \\ 3x^2-2x^2 +4x+7x+12-5 &=& 0 \\ \mathbf{ x^2+11x+7 } &=& \mathbf{0} \\ \hline \end{array}\)

 

Vieta: \(rs=7,~ -(r+s) = 11\)

\(\begin{array}{|rcll|} \hline -(r+s) &=& 11 \\ r+s &=& -11 \\ (r+s)^2 &=& (-11)^2 \\ r^2+2rs+ s^2 &=& 121 \quad &|\quad rs &=& 7 \\ r^2 +2*7 + s^2 &=& 121 \\ r^2+s^2 &=& 121 - 14 \\ \mathbf{ r^2+s^2 } &=& \mathbf{ 107 } \\ \hline \end{array}\)

 

laugh

 May 24, 2021

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