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Math Help - Algebra

  1. #1
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    Algebra

    Given that the roots of the equation x^2 + ax + (a + 2) = 0
    differ by 2, find the possible values of the constant a.

    ans= -2 and 6

    Can anyone show me how to work out this question please?

    cheers
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  2. #2
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    so the roots are
    x = \beta
    x = \beta + 2

    so your factorised expression is (x- \beta)(x - (\beta +2))

    so (x- \beta)(x - (\beta +2)) = x^2 + ax + (a + 2)

    expand giving x^2 + +x( -2 \beta - 2) + \beta (\beta +2) = x^2 + ax + (a + 2)

    compare terms of same order

    giving -2 \beta - 2 = a
    and \beta (\beta +2) = a + 2

    so all you need to do is solve this system to find the two values of a.


    Note an alternative method is to use the quad formula.
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  3. #3
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    Quote Originally Posted by sparky69er View Post
    Given that the roots of the equation x^2 + ax + (a + 2) = 0
    differ by 2, find the possible values of the constant a.

    ans= -2 and 6

    Can anyone show me how to work out this question please?

    cheers
    A slightly different way.

    We know that
    r_1 = \frac{-a}{2} + \sqrt{a^2 - 4(a + 2)}
    and
    r_2 = \frac{-a}{2} - \sqrt{a^2 - 4(a + 2)}
    by the quadratic formula, where r1 and r2 are the solutions to the quadratic.

    Thus
    r_1 - r_2 = \sqrt{a^2 - 4(a + 2)} = 2

    Solve for a.

    -Dan
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