Question

Hello! Please help me work through and solve this problem (it requires the quadratic equation): (3x + 1) ^2 = -2x

Answer 1

Do I need to expand the binomial first or does this problem require a different course?

Answer 2

yes

Answer 3

Okay, so that would get me to 9x^2 + 6x + 2 = -2x
I think

Answer 4

\[(3x+1)^2=(3x)^2+2(3x)(1)+1^2 \\ (3x+1)^2=9x^2+6x+1\]

Answer 5

so should be:
\[9x^2+6x+1=-2x \\ \text{ then add } 2x \text{ on both sides } \\ 9x^2+6x+2x+1=0 \\ 9x^2+8x+1=0\]

Answer 6

Oh yes, sorry. Error on my part! And from here should these numbers be my values for a, b, and c?

Answer 7

yes you can use quadratic formula if you want
a=9
b=8
c=1

in the end you should check your answers
x being positive will not work just so you know
because a real number squared will result in a positive number (or zero)

(3x+1)^2=-2x
we need x to be negative because 0 obviously doesn't work either

Answer 8

and after solving these equation I can tell you you will not have any domain issues with the answers you get

Answer 9

Perfect! I ended up getting -8 +- radical 28 / 18

Answer 10

I guess I would need to simplify this further...

Answer 11

\[x=\frac{-8 \pm \sqrt{28}}{18} \\ \text{ note : } \sqrt{28}=\sqrt{4 \cdot 7} =\sqrt{4} \sqrt{7}=2 \sqrt{7} \\ \]
yes your solution can be simplified
see the note as a hint

Answer 12

Oh, so my final answer would be
x = -4/9 +- radical 7 / 9

Answer 13

\[x=\frac{-8 \pm 2 \sqrt{7}}{18} \\ \text{ divide \top and bottom by 2 } \\ x=\frac{\frac{-8}{2} \pm \frac{2}{2} \sqrt{7}}{\frac{18}{2}} \\ x=\frac{-4 \pm \sqrt{7}}{9} \\ \text{ yes this could be written as } x=\frac{-4}{9} \pm \frac{\sqrt{7}}{9}\]

Answer 14

Thank you!!! This helped a bunch.

Answer 15

cool stuff! :)