Results 1 to 3 of 3

Math Help - No real roots?

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1

    No real roots?

    Determine the possible integer values of n for which there exists some real number k such that the equation 2x^{n}-4x+1=k has no real roots.

    Anyone can help? Thanks!!
    Last edited by acc100jt; September 14th 2009 at 10:32 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    For n = 2 use the discriminant <0.

    2x^{2}-4x+1-k=0

    Solve (-4)^2-4\times 2\times (1-k) < 0

    Maybe this idea is a little simplistic for these problem
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1
    obviously n has to be even, because i need the signs of the limits of the quadratic f(x)=2x^{n}-4x+1 to be the same, when x tend to both +ve and -ve infinity. Then the graph of f(x) will curve either upwards or downwards at both ends, so there will exists some k where the equation has no solution. But how to give a formal solution??
    or is there any other way to solve it???
    Last edited by acc100jt; September 14th 2009 at 10:19 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Real and non-real roots to a quintic equation
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: February 14th 2011, 09:42 PM
  2. Replies: 7
    Last Post: February 10th 2011, 09:34 PM
  3. No. of real roots.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 8th 2011, 08:12 PM
  4. Replies: 8
    Last Post: April 7th 2009, 01:15 PM
  5. Real roots
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 25th 2008, 03:31 PM

Search Tags


/mathhelpforum @mathhelpforum