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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.

 

Thank you in advance

 Oct 12, 2020
 #1
avatar+10319 
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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

laugh  !

 Oct 12, 2020
edited by asinus  Oct 12, 2020
edited by asinus  Oct 12, 2020
 #2
avatar+19 
0

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
avatar+111127 
0

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

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