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Math Help - Quadratic equation help!!!

  1. #1
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    Post Quadratic equation help!!!

    If kx^2-x-k-1=0 calculate the values of k where the roots of the quadratic are:
    a.Real and uneven
    b.Real and have the opposite sign..

    So far i got x=-1 or x =1+1/k for the roots...how do i assess a and b?
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  2. #2
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    Re: Quadratic equation help!!!

    Quote Originally Posted by yeatch View Post
    If kx^2-x-k-1=0 calculate the values of k where the roots of the quadratic are:
    a.Real and uneven
    b.Real and have the opposite sign..

    So far i got x=-1 or x =1+1/k for the roots...how do i assess a and b?
    $x = \dfrac{-\ (-\ 1) \pm \sqrt{(-1)^2 - 4k(-k - 1}}{2k} = \dfrac{1 \pm \sqrt{4k^2 + 4k + 1}}{2k} = \dfrac{1 \pm \sqrt{(2k + 1)^2}}{2k} \implies$

    $x = \dfrac{-2k}{2k} = - 1\ or\ x = \dfrac{1 + 2k + 1}{2k} = \dfrac{2k + 2}{2k} = 1 + \dfrac{1}{k}.$

    Well done so far.

    Notice that the question asks about values (plural) of k. "Uneven" is a bit weird. Normally, one would say odd, which would imply an integer. If "uneven" means "not an even integer," what values of k are acceptable. If "uneven" means an odd integer, what values of k are acceptable. As for part b, notice that one root is negative, what does that imply about k if the other root is positive?
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  3. #3
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    Re: Quadratic equation help!!!

    The integer replies that it should be a whole number not a fraction. I do not see any solutions to this problem...
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  4. #4
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    Re: Quadratic equation help!!!

    do i just chose any value for k as long as its uneven and odd ?
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  5. #5
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    Re: Quadratic equation help!!!

    Quote Originally Posted by yeatch View Post
    do i just chose any value for k as long as its uneven and odd ?
    The condition imposed by the question involves the roots of the equation, not k itself. It is provable that no odd integer can be the root of that equation.

    $Assume\ 1 + \dfrac{1}{k} = 2j + 1,\ where\ j \in \mathbb Z.$

    $0 = x^2 - x - k - 1 = (x + 1)(x - 2j - 1) = x^2 + (1 - 2j - 1)x - 2j - 1 = x^2 - 2jx - 2j - 1 \implies - 2j = - 1 \implies j \not \in \mathbb Z.$

    So, assuming you have transcribed the question correctly, there is NO VALUE of k that will make both roots of that equation odd integers.

    But that still leaves the issue of whether there ARE values of k that will make one root positive and one root negative. Because one root is minus 1, this question boils down to what values of k will make the other root positive.

    If you are still lost, please ask more questions.
    Last edited by JeffM; May 10th 2014 at 08:11 PM. Reason: I left some scratch work in original post
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  6. #6
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    Re: Quadratic equation help!!!

    Hey Jeff, ye olde mudder you; how you doin? Miss Lookagain?
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  7. #7
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    Re: Quadratic equation help!!!

    Quote Originally Posted by Wilmer View Post
    Hey Jeff, ye olde mudder you; how you doin? Miss Lookagain?
    Mon ami, je ne regrette point l'absence de mademoiselle lookagain. Nor mmmbot.
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  8. #8
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    Re: Quadratic equation help!!!

    Quote Originally Posted by JeffM View Post
    Mon ami, je ne regrette point l'absence de mademoiselle lookagain. Nor mmmbot.
    mademoiselle lookagain: LOVE IT!!
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