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Math Help - [SOLVED] solve the equation

  1. #1
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    [SOLVED] solve the equation

    Can someone show me how to solve this equation please? This is part of something bigger, but what I'm stuck on is just algebra
    2x^2+x-1=0

    I can distribute it to x(2x+1) = 1, but am not really sure what to do from there, or if that's even the proper way to go. I would normally try to factor, but I don't think this is factorable. Any assistance is appreciated.
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  2. #2
    Super Member Quacky's Avatar
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    I think it does factorize to give (2x-1)(x+1)=0
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  3. #3
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    you're right, it does factor. I was totally unable to figure that out. Thank you for the assistance.
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  4. #4
    Super Member Quacky's Avatar
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    There is a test you can use to see if something can be factorized:

    For any quadratic, ax^2+bx+c=0,
    Take the discriminant b^2-4(a)(c)
    If the result is a perfect square (i.e. it can be square-rooted and give an integer result), then the sum in question can be factorized.

    So for your example,
    b^2-4(a)(c)
    1^2-4(2)(-1)
    =1+8
    =9
    This is a perfect square so the quadratic can be factorized.
    Last edited by Quacky; February 4th 2010 at 10:58 AM.
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  5. #5
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    Are You sure, You want solve : 2x^2+x-1=0 ??

    So, 2x^2+x-1=0 is a equation the second degré

    \Delta = b^2-4ac = 1^2 - 4*2*(-1) = 9 \geq 0

    The Solutions : x_1= \frac{-1+ \sqrt{9}}{4} = \frac{1}{2} , x_2 = \frac{-1-\sqrt{9}}{4} = -1
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  6. #6
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    Thanks Quacky, that's very instructive. I had no idea. That should come in handy.

    For the record, the equation above is actually the denominator of a larger equation that I was trying to find the vertical asymptotes for. I know to find those, you just set the denominator to 0, hence the original question. I appreciate all the responses.
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