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Math Help - Quadratic Proof

  1. #1
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    Lightbulb Quadratic Proof

    I've been having some trouble with this one:

    Prove that if x^2 > k(x - 2) for all real x, then 0 < k < 8.

    I started with taking the k(x-2) to the other side, but I don't think I'm going anywhere with this.

    Thanks in advance.
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  2. #2
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    Re: Quadratic Proof

    \displaystyle \begin{align*} x^2 &> k(x - 2) \\ x^2 &> kx - 2k \\ x^2 - kx + 2k &> 0\end{align*}

    For this inequation to be true, then the discriminant of \displaystyle x^2 - kx + 2k must be negative.

    So
    \displaystyle \begin{align*} (-k)^2 - 4(1)(2k) &< 0 \\ k^2 - 8k &< 0 \\ k^2 - 8k + (-4)^2 &< (-4)^2 \\ (k - 4)^2 &< 16 \\ |k - 4| &< 4 \\ -4 < k - 4 &< 4 \\ 0 < k &< 8  \end{align*}
    Last edited by Prove It; June 23rd 2011 at 07:45 AM.
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  3. #3
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    Re: Quadratic Proof

    x^2-kx+2k>0

    is true for all x, if the U-shaped quadratic never touches the x-axis,
    so the equation

    x^2-kx+2k=0

    is of the form

    ax^2+bx+c=0

    and must have complex roots

    \Rightarrow\ b^2-4ac<0\Rightarrow\ k^2-8k<0

    \Rightarrow\ k(k-8)<0

    One factor is positive and the other is negative.
    k>8 and k<0 cannot occur, so k>0 and k<8.
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