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# need help

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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   $$r^2+s^2.$$

Hello Guest!

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

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

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

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

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

$$r^2+s^2=15.25$$

!

Oct 25, 2021