Results 1 to 2 of 2

Math Help - Express q as a function...

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    75

    Express q as a function...

    Express q as a function of p, given that one root of x^2 + px + q = 0 is twice the other.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,925
    Thanks
    332
    Awards
    1
    Quote Originally Posted by ceasar_19134 View Post
    Express q as a function of p, given that one root of x^2 + px + q = 0 is twice the other.
    First find the roots:
    x = [-p (+/-) sqrt{p^2 - 4q}]/2

    So we know that x(+) = 2*x(-) (since x(+) must be the larger of the two).

    So:
    [-p + sqrt{p^2 - 4q}]/2 = 2*[-p - sqrt{p^2 - 4q}]/2

    -p + sqrt{p^2 - 4q} = 2*[-p - sqrt{p^2 - 4q}]

    -p + sqrt{p^2 - 4q} = -2*p - 2*sqrt{p^2 - 4q}

    3*sqrt{p^2 - 4q} = -p <-- Square both sides.

    9*(p^2 - 4q) = p^2

    8p^2 - 36q = 0

    2p^2 - 9q = 0

    q = (2/9)p^2

    I'll leave it to you to check that this answer is correct. (Squaring both sides of an equation occasionally introduces "solutions" that aren't correct, so you always need to check you answer.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Express as a Function.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 27th 2011, 09:54 PM
  2. Replies: 2
    Last Post: March 20th 2011, 01:29 AM
  3. Replies: 3
    Last Post: June 4th 2010, 10:40 PM
  4. Express f(g) as a function of x
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 7th 2010, 02:54 PM
  5. Express the function...
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: August 29th 2008, 02:17 PM

Search Tags


/mathhelpforum @mathhelpforum