sighn0more one year ago Hello! Please help me work through and solve this problem (it requires the quadratic equation): (3x + 1) ^2 = -2x

1. sighn0more

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

2. freckles

yes

3. sighn0more

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

4. freckles

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

5. freckles

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

6. sighn0more

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

7. freckles

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

8. freckles

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

9. sighn0more

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

10. sighn0more

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

11. freckles

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

12. sighn0more

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

13. freckles

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

14. sighn0more

Thank you!!! This helped a bunch.

15. freckles

cool stuff! :)