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

 Aug 4, 2020
 #1
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\(x = {-4 \pm \sqrt{4^2-4(3)(12)} \over 2(3)}\\ x = {-4 \pm \sqrt{-128} \over 6}\\ x = {-4 \pm 3\sqrt{-2} \over 6}\\ ({-4 + 3\sqrt{-2} \over 6})^2+({-4 - 3\sqrt{-2} \over 6})^2\\ {(-4 + 3\sqrt{-2})^2 \over 6^2}+{(-4 - 3\sqrt{-2})^2 \over 6^2}\\ {(-4 + 3\sqrt{-2})^2 \over 36}+{(-4 - 3\sqrt{-2})^2 \over 36}\\ {(-4 + 3\sqrt{-2})^2+(-4 - 3\sqrt{-2})^2 \over 36}\\\)

Well this is kind of bashy, lOl.

\(r + s = -\frac{4}{3} \text{ due to vieta}\\ (r + s)^2 = \frac{16}{9} \text{ because we square -4}\\ r^2+s^2+2rs=\frac{16}9\\ rs=4 \text{ due to vieta}\\ r^2 + s^2 = \frac{16}{9} - 2(4) = \frac{16}{9}-8=-6\frac{2}{9}\)

Someone check this

 Aug 4, 2020

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