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.
explanation: The document is a conversation between a student and others, primarily discussing solving quadratic and cubic equations. The student initially presents several equations and asks for help, and the others provide guidance and examples. The document contains multiple questions and answers related to solving these equations.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a conversation between a student and others, primarily discussing solving quadratic and cubic equations. The student initially presents several equations and asks for help, and the others provide guidance and examples. The document contains multiple questions and answers related to solving these equations.
  
  Label: multiple-question-and-answer