text: 1. ## pre-calc stress
  
  I've been staring at the following problems for at least half an hour. Would anyone help me?
  
  2x^2+8x=0
  -9x^2+12x-4=0
  4x^2-4x+2=0
  3x^2+4x-1=0
  x^3+2x^2-4x-8=0
  
  2. Do you know how to use the quad formula?
  
  $\frac{-b_-^+ \sqrt{b^2-4ac}}{2a}$
  
  3. I've tried, but I keep getting answers I'm sure are way off from being right.
  
  4. What kind of answers do you keep getting?
  
  5. for the first one i got .396 and 8.48
  
  6. I'm going to do one for you for an example
  
  $-9x^2+12x-4=0$
  
  $Ax^2+Bx+C=0$
  
  $\frac{-12_-^+ \sqrt{144-4(-9)(-4)}}{-18}$
  
  $\frac{-12_+^- 0}{-18}$
  
  $\frac{1}{3}$
  
  7. Thank you so much. I think I have an idea of how to do the rest.
  
  8. Ok, I got stuck on x^3+2x^2-4x-8=0
  
  9. Originally Posted by 11rdc11
  I'm going to do one for you for an example
  
  $-9x^2+12x-4=0$
  
  $Ax^2+Bx+C=0$
  
  $\frac{-12_-^+ \sqrt{144-4(-9)(-4)}}{-18}$
  
  $\frac{-12_+^- 0}{-18}$
  
  $\frac{1}{3}$
  Yo
  
  Next time, you can use $\pm$ (\pm) and $\mp$ (\mp)
  
  10. Originally Posted by nina9
  Ok, I got stuck on x^3+2x^2-4x-8=0
  Try x - 2 as a factor. $x^3 + 2x^2 - 4x - 8 = (x - 2)(x^2 + 4x + 4)$ using long or synthetic division.
  
  11. For 2x^2+8x=0, both the terms have x in them so this can be factorised out:
  
  x(2x+8)=0
  
  I was going to do the rest of them by finding a factor, but they all appear to be decimals or surds apart from the last one. I'll show you the method so you can use it on other questions:
  
  f(x)=x^3+2x^2-4x-8
  
  f(2)=2^3+2^3-8-8
  f(2)=8+8-8-8
  f(2)=0
  
  Since f(2)=0, f(x) has a factor x-2 ie. when x=2 f(x)=0.
  
  This can then be factorised using this (see icemanfan's post).
  
  A graphical method can help:
  
  If in doubt, draw the graph and see if that helps.