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Math Help - [SOLVED] Show the roots of this equation : how?

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    [SOLVED] Show the roots of this equation : how?

    Show that the roots of this quadratic equation (k + 5)x^2 + 13x -6k = 0 are real and distinct for all the real values of k.

    This was how I did it.

    a = (K + 5), b = 13, c = -6k

    Using b - 4ac,

     (13) - 4(k + 5)(-6k)
     = 169 + 24k + 120k
     = 24k^2 + 120k + 169

    Using the formula of finding the roots of a quadratic equation,

     \frac{-120 \pm \sqrt{(120)^2 - 4(24)(169)}}{2(24)}

     \frac{-120 \pm \sqrt{-1824}}{2(24)}

    After this, I don't know what to do. I know that this way is wrong so please correct me on the steps.
    Last edited by mark1950; July 7th 2009 at 04:23 AM.
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  2. #2
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    Quote Originally Posted by mark1950 View Post
    Show that the roots of this quadratic equation (k + 5)x^2 + 13x -6k = 0 are real and distinct for all the real root values of k.

    ...
    Solve this equation for x using the quadratic formula:

    x=\dfrac{-13 \pm \sqrt{169-4(k+5)\cdot(-6k)}}{2(k+5)}

    The roots are real if the radical is positive (or zero) for all values of k:

    x=\dfrac{-13 \pm \sqrt{169+24k^2+120k}}{2(k+5)}

    x=\dfrac{-13 \pm \sqrt{24(k^2+5k + \dfrac{25}{4})-\dfrac{24\cdot25}{4}+169}}{2(k+5)}

    x=\dfrac{-13 \pm \sqrt{24(k^2+\dfrac52)^2+19}}{2(k+5)}

    Thus the radical is at least +19 and therefore there must exist real roots of the equation.
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  3. #3
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    Quote Originally Posted by mark1950 View Post
    Show that the roots of this quadratic equation (k + 5)x^2 + 13x -6k = 0 are real and distinct for all the real root values of k.
    I'm not sure that the bolded word should be there.

    This was how I did it.

    a = (K + 5), b = 13, c = -6k

    Using b - 4ac,

     (13) - 4(k + 5)(-6k)
     = 169 + 24k + 120k
     = 24k^2 + 120k + 169

    Using the formula of finding the roots of a quadratic equation,

     \frac{-120 \pm \sqrt{(120)^2 - 4(24)(169)}}{2(24)}
    If the roots of (k + 5)x^2 + 13x -6k = 0 are real and distinct, then it should be 24k^2 + 120k + 169 {\color{red}>} 0, not 24k^2 + 120k + 169 {\color{red}=} 0.

    When solving a quadratic inequality, the first thing I do is change it to an equation anyway to find the critical points. So assuming that you plugged in numbers into the quadratic formula right:
     \frac{-120 \pm \sqrt{-1824}}{2(24)}

    This tells you that the quadratic 24k^2 + 120k + 169 = 0 has no real roots, that it doesn't cross the x-axis. That means that for any value of k, the expression 24k^2 + 120k + 169 will always be positive or always be negative. Since the leading coefficient is positive, 24k^2 + 120k + 169 > 0 for any k. Therefore, the roots of (k + 5)x^2 + 13x -6k = 0 are real and distinct.


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    oh.......I get it now. Lol. Thanks, man!
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  5. #5
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    Quote Originally Posted by mark1950 View Post
    For all values of k, 24k + 120k + 169 > 0. This, I now know. But how does this nature of k influence the roots of the quadratic equation (k + 5)x + 13x - 6k = 0 to be real and distinct? Examples would be helpful. Thanks.
    The discriminant of the quadratic equation
    (k + 5)x^2 + 13x - 6k = 0,
    as you have shown, is
    24k^2 + 120k + 169.

    Recall that for the discriminant b - 4ac:
    if b - 4ac > 0, there are two distinct real roots,
    if b - 4ac = 0, there is one real double root, and
    if b - 4ac < 0, there are no real roots.

    Since we have shown that 24k^2 + 120k + 169 > 0 for any real number k, there will be two distinct real roots in the quadratic (k + 5)x^2 + 13x - 6k = 0, for any real number k.


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