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Math Help - Quadratic equation

  1. #1
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    Quadratic equation

    Hey!

    Someone please answer the following question, I tried everything but I still couldn't do it right...

    Given that the equation x^2+ax=b, where a and b are real numbers, has a unique solution, prove that a^2+4b=0

    I would very much appreciate it if someone can show me how it is done.

    Thanks in advance
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  2. #2
    Super Member Quacky's Avatar
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    Re: Quadratic equation

    x^2+ax-b=0

    In order for there to be 1 repeated root, the discriminant of the quadratic has to be 0.

    Can you remember how to calculate the discriminant?
    Last edited by Quacky; October 16th 2011 at 06:04 AM.
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  3. #3
    MHF Contributor

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    Re: Quadratic equation

    an alternate approach:

    given that we have just one solution:

    x^2 + ax - b = (x - r)^2 = 0

    multiplying the square out, we get:

    x^2 + ax - b = x^2 - 2rx + r^2

    thus  a = -2r, b = -r^2.

    now, we want to end up with a condition on a and b, but we don't know what they are. but if we work in terms of r, maybe we can come up with something.

    well b is in terms of r^2, but a is only in terms of r. so let's square a:

    a^2 = (-2r)^2 = 4r^2.

    that's almost the "opposite of b" we just need to multiply b by 4:

    4b = -4r^2.

    now, r only exists (and the equations for a and b involving r), because we are assuming we have just one solution.

    but IF WE DO, then it is clear that:

    a^2 + 4b = 4r^2 - 4r^2 = 0

    (and we didn't even have to remember what the discriminant is, or use that horrid quadratic formula)
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