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Math Help - Equal Real Roots

  1. #1
    Junior Member casey_k's Avatar
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    Equal Real Roots

    f(x) = -1/2x^2 -5x +4

    find the value(s) of k such that the equation f(x+k)=0 has two equal real roots.
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  2. #2
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    Quote Originally Posted by casey_k View Post
    f(x) = -1/2x^2 -5x +4

    find the value(s) of k such that the equation f(x+k)=0 has two equal real roots.
     f(x+k) = \frac{-1}{2}(x+k)^2 -5(x+k) + 4

     f(x+k) = \frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4

     f(x+k) = \frac{-1}{2}x^2-\frac{1}{2}2kx-\frac{1}{2}k^2 -5x-5k + 4

     f(x+k) = \frac{-1}{2}x^2+x(-k-5) - \frac{k^2}{2} -5k + 4

    The discriminant must be zero for equal roots:

     b^2-4ac = 0

     a = \frac{-1}{2}

     b = (-k-5)

     c = - \frac{k^2}{2} -5k + 4

    Insert these into the equation. Expand. Then solve for k.
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  3. #3
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    Quote Originally Posted by Mush View Post
     f(x+k) = \frac{-1}{2}(x+k)^2 -5(x+k) + 4

     f(x+k) = \frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4

     f(x+k) = \frac{-1}{2}x^2-\frac{1}{2}2kx-\frac{1}{2}k^2 -5x-5k + 4

     f(x+k) = \frac{-1}{2}x^2+x(-k-5) - \frac{k^2}{2} -5k + 4

    The discriminant must be zero for equal roots:

    **  b^2-4ac = 0

    **  a = \frac{-1}{2}

    **  b = (-k-5)

    **  c = - \frac{k^2}{2} -5k + 4

    Insert these into the equation. Expand. Then solve for k.
    Problem is that no k will solve this set of equations (**). If you look at the graph of f(x), f(x+k) only translates the graph left and right. There will always be two real roots (never repeated). Now if the question read f(x)+k then we could translate up and down to get a repeated real root.
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