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Thread: disicriment of solutions

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    disicriment of solutions

    $\displaystyle 6x^2+4kx+(k+3)=0$

    i)find one solution
    ii) two
    iii)none
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    Quote Originally Posted by johnsy123 View Post
    $\displaystyle 6x^2+4kx+(k+3)=0$

    i)find one solution
    ii) two
    iii)none
    Compare your equation to the standard form of a quadratic equation.

    Quadratic: $\displaystyle ax^2+bx+c=0$

    Yours: $\displaystyle 6x^2 + 4kx + (k+3) = 0$

    Assuming k is a constant we can match the coefficients:

    • $\displaystyle a = 6$
    • $\displaystyle b = 4k$
    • $\displaystyle c = k+3$


    The discriminant is given by: $\displaystyle \Delta = b^2-4ac$

    If:
    • $\displaystyle \Delta > 0$ then there are two, different, real solutions
    • $\displaystyle \Delta = 0$ then there are two, equal, real solutions
    • $\displaystyle \Delta < 0$ then there are no real solutions


    I will let you figure out the rest of the question
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    Quote Originally Posted by johnsy123 View Post
    $\displaystyle 6x^2+4kx+(k+3)=0$

    i)find one solution
    ii) two
    iii)none
    For all of the above, you need to evaluate the discriminant.

    $\displaystyle \Delta = b^2 - 4ac$

    $\displaystyle = (4k)^2 - 4(6)(k + 3)$

    $\displaystyle = 16k^2 - 24k - 72$.


    For one solution, $\displaystyle \Delta = 0$.

    So $\displaystyle 16k^2 - 24k - 72 = 0$

    $\displaystyle 8(2k^2 - 3k - 9) = 0$

    $\displaystyle 8(2k^2 - 6k + 3k - 9) = 0$

    $\displaystyle 8[2k(k - 3) + 3(k - 3)] = 0$

    $\displaystyle 8(k - 3)(2k + 3)= 0$.


    So $\displaystyle k - 3 = 0$ or $\displaystyle 2k + 3 = 0$

    $\displaystyle k = 3$ or $\displaystyle k = -\frac{3}{2}$, for one solution.



    No solutions: $\displaystyle \Delta < 0$.

    So $\displaystyle 16k^2 - 24k - 72 < 0$.

    To evaluate this, you will need to complete the square.

    $\displaystyle 16\left(k^2 - \frac{3}{2}k - \frac{9}{2}\right) < 0$

    $\displaystyle 16\left[k^2 - \frac{3}{2}k + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 - \frac{9}{2}\right] < 0$

    $\displaystyle 16\left[\left(k - \frac{3}{4}\right)^2 - \frac{9}{16} - \frac{72}{16}\right] < 0$

    $\displaystyle 16\left[\left(k - \frac{3}{4}\right)^2 - \frac{81}{16}\right] < 0$

    $\displaystyle \left(k - \frac{3}{4}\right)^2 - \frac{81}{16} < 0$

    $\displaystyle \left(k - \frac{3}{4}\right)^2 < \frac{81}{16}$

    $\displaystyle \sqrt{\left(k - \frac{3}{4}\right)^2} < \sqrt{\frac{81}{16}}$

    $\displaystyle \left|k - \frac{3}{4}\right| < \frac{9}{4}$

    $\displaystyle -\frac{9}{4} < k - \frac{3}{4} < \frac{9}{4}$

    $\displaystyle -\frac{3}{2} < k < 3$.



    2 solutions: $\displaystyle \Delta > 0 $.

    So $\displaystyle 16k^2 - 24k - 72 > 0$.

    To evaluate this, again you will need to complete the square.

    $\displaystyle 16\left(k^2 - \frac{3}{2}k - \frac{9}{2}\right) > 0$

    $\displaystyle 16\left[k^2 - \frac{3}{2}k + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 - \frac{9}{2}\right] > 0$

    $\displaystyle 16\left[\left(k - \frac{3}{4}\right)^2 - \frac{9}{16} - \frac{72}{16}\right] > 0$

    $\displaystyle 16\left[\left(k - \frac{3}{4}\right)^2 - \frac{81}{16}\right] > 0$

    $\displaystyle \left(k - \frac{3}{4}\right)^2 - \frac{81}{16} > 0$

    $\displaystyle \left(k - \frac{3}{4}\right)^2 > \frac{81}{16}$

    $\displaystyle \sqrt{\left(k - \frac{3}{4}\right)^2} > \sqrt{\frac{81}{16}}$

    $\displaystyle \left|k - \frac{3}{4}\right| > \frac{9}{4}$

    $\displaystyle k - \frac{3}{4} < -\frac{9}{4}$ or $\displaystyle k - \frac{3}{4} > \frac{9}{4}$

    $\displaystyle k < -\frac{3}{2}$ or $\displaystyle k > 3$.
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