Results 1 to 3 of 3

Math Help - disicriment of solutions

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    108

    disicriment of solutions

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

    i)find one solution
    ii) two
    iii)none
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by johnsy123 View Post
    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: ax^2+bx+c=0

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

    Assuming k is a constant we can match the coefficients:

    • a = 6
    • b = 4k
    • c = k+3


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

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


    I will let you figure out the rest of the question
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,319
    Thanks
    1238
    Quote Originally Posted by johnsy123 View Post
    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.

    \Delta = b^2 - 4ac

     = (4k)^2 - 4(6)(k + 3)

     = 16k^2 - 24k - 72.


    For one solution, \Delta = 0.

    So 16k^2 - 24k - 72 = 0

    8(2k^2 - 3k - 9) = 0

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

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

    8(k - 3)(2k + 3)= 0.


    So k - 3 = 0 or 2k + 3 = 0

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



    No solutions: \Delta < 0.

    So 16k^2 - 24k - 72 < 0.

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

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

    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

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

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

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

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

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

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

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

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



    2 solutions: \Delta > 0 .

    So 16k^2 - 24k - 72 > 0.

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

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

    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

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

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

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

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

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

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

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

    k < -\frac{3}{2} or k > 3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 4th 2011, 08:21 PM
  2. Replies: 6
    Last Post: July 26th 2010, 11:45 AM
  3. Solutions
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 25th 2010, 11:01 AM
  4. solutions..
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 30th 2010, 01:02 AM
  5. All non-solutions?
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 5th 2009, 04:13 PM

Search Tags


/mathhelpforum @mathhelpforum