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Quadratic equation

  1. Feb 11, 2014 #1
    1.given the equation a(3x2+2x+1)=4x-6x2-4 with the solutions x1 and x2
    a) let a=0 without solving the equation calculate x1-3+x2-3

    the correct answer is supposed to be -7/2

    2. Relevant equations

    3. The attempt at a solution
    The first thing I was that I put in "a" so,
    0=4x-6x2-4................... here I multiplied the whole equation by -1
    0=6x2-4x+4...........................here i divided the equation by 6
    then I used the equation
    1/x13+1/x23=(1/x1+1/x2)3 -3/x12x2 -3/x22x1
    This is where I have no idea how to continue.I'm not even sure that what I have written is correct .If someone could help me solve this,I would really appreciate it. Thanks for reading .
  2. jcsd
  3. Feb 11, 2014 #2


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    Try to bring in x1+x2 and x1x2.
    For example, [tex]1/x_1+1/x_2=\frac{x_1+x_2}{x_1 x_2}[/tex]

  4. Feb 11, 2014 #3
    thank you for your reply. I did as you said and I got:
    ((x1+x2)/x1x2)3-3x1/x12x22 -3x2/x22x12. then I added the minuses together:
    Did I do this part correct?
    May I also ask, what command did you use to write (x1+x2)/x1x2 because your way looks much easier to understand
    Last edited: Feb 11, 2014
  5. Feb 11, 2014 #4

    Ray Vickson

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    He did not write ##x_1+x_2/x_1x_2##; you did that. In fact, you wrote
    [tex]x_1 + \frac{x_2}{x_1 x_2}[/tex]
    If you mean
    [tex] \frac{x_1 + x_2}{x_1 x_2}[/tex]
    then either use LaTeX or else use parentheses, like this: (x1+x2)/x1x2.
  6. Feb 11, 2014 #5
    Thank you I didn't notice that and I fixed it right away
  7. Feb 11, 2014 #6
    Now I added in the numbers
    (((4/6)/4/6))-((3*4/6)/(4/6)2 =
    Thank you very much for your help I got the right answer
  8. Feb 11, 2014 #7


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