Results 1 to 3 of 3

Math Help - equations

  1. #1
    Member Veronica1999's Avatar
    Joined
    Jan 2010
    From
    Jupimar
    Posts
    160

    equations

    For one root of ax(squared) + bx + c to be double the other, the coefficients a, b, c, must be related as follows:

    a. 4b(squared) = 9c
    b. 2b(squared) = 9ac
    c. 2b(squared) = 9a
    d. b(squared) - 8ac = 0
    e. 9b(squared) = 2ac

    I got the answer b by using (x-2)(x-4)=0.
    I plugged in a=1 b= -6 c=8 but I think the problem wasn't meant to be solved that way. Is there a better way to do it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2009
    From
    New Delhi
    Posts
    153
    For one root of  ax^2 + bx + c to be double the other

    then

     \frac {-b- \sqrt{b^2-4ac} }{2a} = 2 \times  \frac {-b+ \sqrt{b^2-4ac} }{2a}
    solving above equation we get
     2b^2=9ac
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Another solution...

    ax^2+bx+c

    if x_1 and x_2 are two roots, we demand on x_2 to be x_2=2x_1

    Now, sum of the two roots is (*) x_1+x_2=3x_1=\frac{-b}{a} and their product is x_1x_2=2x_1^2=\frac{c}{a}

    So, x_1=\sqrt{\frac{c}{2a}} put this to (*), and you get what you need to prove.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 02:41 AM
  2. Replies: 3
    Last Post: February 27th 2009, 08:05 PM
  3. Replies: 3
    Last Post: December 2nd 2008, 11:54 AM
  4. Replies: 1
    Last Post: September 1st 2007, 07:35 AM
  5. Replies: 1
    Last Post: July 29th 2007, 03:37 PM

Search Tags


/mathhelpforum @mathhelpforum