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Math Help - Roots of a Quadraitc Equation

  1. #1
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    Roots of a Quadraitc Equation

    Hi,
    I would appreciate any help with the following question.

    Given that the roots of the equation x + ax + (a + 2) = 0 differ by two, find the possible values of constant a.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Roots of a Quadraitc Equation

    Use the quadratic equation to get an expression for the two roots of the equation, and set the difference between them equal to two:

    x_1 = \frac {-a + \sqrt{a^2-4(a+2)}}2

    x_2 = \frac {-a - \sqrt{a^2-4(a+2)}}2

    subtract, and set equal to 2:

    x_1 - x_2 = \sqrt{a^2-4(a+2)} = 2

    Now solve for a - square both sides, rearrange, and use the quadratic formula.
    Last edited by ebaines; July 3rd 2013 at 10:32 AM.
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  3. #3
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    Re: Roots of a Quadraitc Equation

    Another way: (x- x_0)(x- x_1)= x^2- (x_0+ x_1)x+ x_0x_1= 0 so if x_0 and x_1 are roots of x^2+ ax+ (a+ 2)= 0 then we must have x_0+ x_1= a and x_0x_1= a+ 2. Knowing also that "the roots of the equation x + ax + (a + 2) = 0 differ by two", so that x_1= x_0+ 2, we have three equations to solve for the three values, x_0, x_1, a.
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  4. #4
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    Re: Roots of a Quadraitc Equation

    Hi,
    Thanks very much for your help.
    Much appreciated.
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  5. #5
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    Re: Roots of a Quadraitc Equation

    Thanks a lot for your help.
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