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

 Oct 25, 2021
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Find   \(r^2+s^2. \)

 

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\(3x^2+4x+12 = x^2 - 5x + 7\\ 2x^2+9x+5=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-9 \pm \sqrt{9^2-4\cdot 2\cdot 5} \over 2\cdot 2}\\ x=\frac{-9\pm 6.4031}{4}\)

\(r=\frac{-9+\sqrt{41 } }{4}\\ r^2=\frac{81-18\sqrt{41 }+41}{16}\\ s=\frac{-9-\sqrt{41}}{4}\\ s^2=\frac{81+18\sqrt{41 }+41}{16}\\ \)

\(r^2+s^2=\frac{2\cdot(81+41)}{16}\)

\(r^2+s^2=15.25\)

laugh  !

 Oct 25, 2021

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