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Math Help - simple quadratics problem

  1. #1
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    simple quadratics problem

    if given the eqn:
    (=squared)
    -x +5x +2=0

    then i do the add/mult thing to find the roots, so we have to find two factors which add to give +5 and multiply to give...?
    do i always multiply the +2 (c) by the coefficient of the X (a)?

    therefore we have to find something which multiplies to give -2?
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  2. #2
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    The equation is

    -x^{2}+5x+2=0, which I would probably write

    x^{2}-5x-2=0, so as to get the coefficient of the highest power of x to be one. I don't think factoring is going to be the best way to solve this problem. You'd have to come up with two numbers whose product is -2, and which add up to -5. You could, of course, set up two simultaneous equations:

    uv=-2
    u+v=-5, and solve. But that's adding another quadratic equation to solve, which seems a bit silly.

    Have you done completing the square, or the quadratic formula yet?
    Last edited by Ackbeet; October 28th 2010 at 07:14 AM. Reason: One versus zero.
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  3. #3
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    Quote Originally Posted by chartsy View Post
    if given the eqn:
    (=squared)
    -x +5x +2=0

    then i do the add/mult thing to find the roots, so we have to find two factors which add to give +5 and multiply to give...?
    do i always multiply the +2 (c) by the coefficient of the X (a)?

    therefore we have to find something which multiplies to give -2?
    first multiply through by negative 1, to make factorising easier,

     x^{2} -5x - 2 = 0

    You have to find two numbers that multiply to give you -2 and add up to give -5.

    so you would have to use the quadratic equation to solve this one.

     x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

    a = 1 , b = -5 and c = -2, put these values into the equation above and you should get two values of x,
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  4. #4
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    lol actually i had a different forumla in my question , idon't have it by me, but can you just tell me if by factoring, the right thing to do is multiply c by the coefficient of X, even if it is negative?
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  5. #5
    A Plied Mathematician
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    Well, you can always multiply the entire equation by a nonzero constant without changing the solutions. You can't JUST multiply c by the coefficient of x.
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