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Im trying to find the two values of X

  1. Nov 2, 2004 #1

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    I have this question and I want to know if its correct. Im using the fomula X=-b +/- {root sign} b - 4ac all over 2a

    This is the question 0=6x^2 + X + -1
    a=6
    b=1
    c=-1

    x=-6 +/- {root sign} 1 - 4(6)(-1) all over 12

    x=-6 +/- root 25 all over 12

    x=-6 +/- 5 all over 12

    Is this the final answer? Did I make a mistake? Im trying to find the two values of X
     
  2. jcsd
  3. Nov 2, 2004 #2
    it is indeed:
    [tex]x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}[/tex]
    [tex]x_{2} = \frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]

    for ax²+bx+c = 0

    marlon, so it looks ok to me
     
  4. Nov 2, 2004 #3

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    cool thanks, that was quick
     
  5. Nov 2, 2004 #4
    as always....

    marlonissimo
     
  6. Nov 2, 2004 #5
    Try to see if you can simplify it down further. -6 + 5 = -1 so one of the answers is -1/12.
     
  7. Nov 2, 2004 #6
    it should be: x=-1 +/- 5 all over 12

    You can of course always check whether your answer is right by using the answer in the original equation and seeing whether it really gives zero, so
    x1 = 4/12
    x2 = -6/12
    and check whether using x1 and x2 in 6x^2 + X + -1 gives 0
     
  8. Nov 2, 2004 #7
    That is right. I didn't check that.

    But, that way you can still simplify it further x1 = 1 / 3 and x2 = -1 / 2.
     
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