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# Algebra Help! Im stuck

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Hi, using vietas formula, ive gotten -4 and -8, but its wrong, can someone help?

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

Oct 12, 2020

#1
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Let r and s be the roots of 3x^2+4x+12=0. Find r^2+s^2.

Hello Age2!

$$3x^2+4x+12=0$$

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

$$x = {-2 \pm \sqrt{16-4\cdot 3\cdot 12} \over 2\cdot 3}$$

The parabola has no real solution.

$$x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{128}\\ x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{2\cdot 2^6}\\ x=-\frac{2}{6}\pm \frac{8i}{6}\cdot \sqrt{2}\\ \color{blue}x=-\frac{1}{3}\pm \frac{4}{3}\cdot \sqrt{2}\cdot i\\$$

$$r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i$$

$$r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i$$

$$r^2=\frac{1}{9}-\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1\\ s^2=\frac{1}{9}+\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1$$

to be continued !

Oct 12, 2020
edited by asinus  Oct 12, 2020
edited by asinus  Oct 12, 2020
#2
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Hello there asinus, thanks for the reply, much appreciated! I see your "to be continued"answers for r squared and s squared, and I;ve been trying to finish it on my own. I just dont know how to add the values together, please help! Thank you

Age26  Oct 12, 2020
#3
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What rubbish, you do not try and do anything on your own.

You just want asinus to finish spoon-feeding you.

Melody  Oct 13, 2020