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Math Help - Quadratic equations: what possible values of k.

  1. #1
    rcs
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    Quadratic equations: what possible values of k.

    How is it possible to solve this ? Determine values(s) of k for which
    1. kx^2 x 4 = 0
    2. 3x^2 + 5x + k = 0
    3. 4x^2 3x + k = 0
    4. kx^2 + 3x - 7 = 0
    5. kx^2 - x - 3 = 0


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    Re: what possible values of k.

    Quote Originally Posted by rcs View Post
    How is it possible to solve this ? Determine values(s) of k for which
    1. kx^2 x 4 = 0
    2. 3x^2 + 5x + k = 0
    3. 4x^2 3x + k = 0
    4. kx^2 + 3x - 7 = 0
    5. kx^2 - x - 3 = 0


    Thank you
    Hi rcs,

    This system is inconsistent. Since,

    kx^{2}-x-4=0\mbox{ and }kx^{2}-x-3=0\Rightarrow 4=3 ; which is obviously a contradiction. Hence the system does not have any solution.
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    Re: what possible values of k.

    Hello, rcs!

    \text{Determine values(s) of }k\text{ for which:}

    . . \begin{array}{ccccc}1.& kx^2  x  4 &=&  0 \\	2.& 3x^2 + 5x + k &=& 0 \\ 3. & 4x^2  3x + k &=& 0  \\ 4.&  kx^2 + 3x - 7 &=& 0 \\ 5. &  kx^2 - x - 3 &=& 0 \end{array}
    .

    The problem is not clearly stated.


    If complex roots are permitted, any value of k can be used.

    If the roots must be real, the discriminant must be nonnegative.

    If the roots must be rational, the discriminant must be a square.

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