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Thread: Quadratic equations nature of roots

  1. #1
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    Quadratic equations nature of roots


    In question number 2,
    I did like this,
    k+4/k−2 >0
    And also i put −b/a value in the equation <0
    Then used quadratic formula
    64−4(k-2)(k+4) <0
    And by trail and error,i got answer as 2.
    Is there another way rather than intuition
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Quadratic equations nature of roots

    We are given:

    $\displaystyle (k-2)x^2+8x+(k+4)=0$

    In order for the roots to be distinct and real, we require the discriminant to be positive:

    $\displaystyle 8^2-4(k-2)(k+4)>0$

    Solving this, we obtain:

    $\displaystyle -6<k<4$

    In order for the roots to be negative, we require:

    $\displaystyle -8+\sqrt{8^2-4(k-2)(k+4)}<0$

    Solving this, we obtain the intervals:

    $\displaystyle [-6,-4)\,\cup\,(2,4]$

    Putting the two together, we obtain:

    $\displaystyle (-6,-4)\,\cup\,(2,4)$

    And so we find (given $\displaystyle k\in\mathbb{Z}$):

    $\displaystyle k\in\{-5,3\}$

    Two values, as you found.
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    Re: Quadratic equations nature of roots

    for $\displaystyle k=-5 $ the roots are positive
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Quadratic equations nature of roots

    Quote Originally Posted by Idea View Post
    for $\displaystyle k=-5 $ the roots are positive
    Yes, in my haste I didn't consider the sign of the denominator of the quadratic formula. So, we need to consider:

    $\displaystyle \frac{-8+\sqrt{8^2-4(k-2)(k+4)}}{k-2}<0$

    Which yields:

    $\displaystyle (-4,2)\,\cup\,(2,4]$

    And then:

    $\displaystyle \frac{-8-\sqrt{8^2-4(k-2)(k+4)}}{k-2}<0$

    Which yields:

    $\displaystyle (2,4]$

    Putting all the restrictions together, we get:

    $\displaystyle (2,4)$

    And so $\displaystyle k=3$ is the only solution satisfying the given conditions.
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