Results 1 to 5 of 5

Math Help - Solve for modulus inequality.

  1. #1
    Junior Member
    Joined
    Aug 2011
    Posts
    43

    Solve for modulus inequality.

    Solve \frac{|x^2-5x+4|}{|x^2-4|}\le1

    as

    |x^2-4|will be positive always

    cross multiply and take 1 to other side of equation
    solve by taking LCM
    we get
    |x^2-5x+4|-(x^2-4)\le0
    on solving we get

    (x^2-5x+4)-(x^2-4)\le0 and -(x^2-5x+4)-(x^2-4)\le0

    the other method I know is to square to remove the modulus function
    (x^2-5x+4)^2-(x^2-4)^2\le0


    among these which method is correct?
    the second method becomes equation of degree 4 i.e. x^4)
    please provide hints.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,795
    Thanks
    1575

    Re: Solve for modulus inequality.

    Quote Originally Posted by sumedh View Post
    Solve \frac{|x^2-5x+4|}{|x^2-4|}\le1

    as

    |x^2-4|will be positive always

    cross multiply and take 1 to other side of equation
    solve by taking LCM
    we get
    |x^2-5x+4|-(x^2-4)\le0
    on solving we get

    (x^2-5x+4)-(x^2-4)\le0 and -(x^2-5x+4)-(x^2-4)\le0

    the other method I know is to square to remove the modulus function
    (x^2-5x+4)^2-(x^2-4)^2\le0


    among these which method is correct?
    the second method becomes equation of degree 4 i.e. x^4)
    please provide hints.
    \displaystyle \begin{align*}\frac{|x^2 - 5x + 4|}{|x^2 - 4|} &\leq 1 \\ |x^2 - 5x + 4| &\leq |x^2 - 4| \\ (x^2 - 5x + 4)^2 &\leq (x^2 - 4)^2 \\ x^4-10x^3+33x^2-40x+16 &\leq x^4 - 8x^2 + 16 \\ 0 &\leq 10x^3 - 41x^2 + 40x \\ 0 &\leq x(10x^2 - 41x + 40) \\ 0 &\leq x\left(10x^2 - 16x - 25x + 40\right) \\ 0 &\leq x\left[2x(5x - 8) - 5(5x - 8)\right] \\ 0 &\leq x(5x - 8)(2x - 5) \end{align*}

    Clearly when \displaystyle f(x) = x(5x - 8)(2x - 5) = 0 we have \displaystyle x = 0 or \displaystyle x = \frac{8}{5} or \displaystyle x = \frac{5}{2}. Since the function changes sign at the x-intercepts, checking the function value at values of x in between the x-intercepts will tell us the intervals where the function is positive.

    \displaystyle f(-1) = -91, so when \displaystyle x < 0 the function is negative.

    \displaystyle f\left(1 \right) = 9, so when \displaystyle 0 < x < \frac{8}{5}, the function is positive.

    \displaystyle f(2) = -4, so when \displaystyle \frac{8}{5} < x < \frac{5}{2}, the function is negative.

    \displaystyle f(3) = 21, so when \displaystyle x > \frac{5}{2}, the function is positive.

    So this inequality is satisfied when \displaystyle x \in \left[0, \frac{8}{5}\right] \cup \left[\frac{5}{2}, \infty\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2011
    Posts
    43

    Re: Solve for modulus inequality.

    Thank you very much
    I was making mistake in solving(X^2-4)
    for checking the sign of the function
    is it easy to put random values before, between and after the zeros to check the sign
    or to make the sign table(attached)??
    Attached Thumbnails Attached Thumbnails Solve for modulus inequality.-untitled.png  
    Last edited by sumedh; August 27th 2011 at 10:14 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,795
    Thanks
    1575

    Re: Solve for modulus inequality.

    Quote Originally Posted by sumedh View Post
    Thank you very much
    I was making mistake in solving(X^2-4)
    for checking the sign of the function
    is it easy to put random values before, between and after the zeros to check the sign
    or to make the sign table(attached)??
    I don't understand what you're doing with this "sign table". It is straightforward enough to substitute values that lie in the required interval to determine the nature of f(x).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2011
    Posts
    43

    Re: Solve for modulus inequality.

    thank you very much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 5th 2011, 11:47 PM
  2. Solving inequality with modulus
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 3rd 2011, 03:37 AM
  3. Solve for x in the inequality
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 3rd 2010, 03:54 AM
  4. modulus inequality can't find similar examples!!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 7th 2008, 04:43 PM
  5. solve the inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 7th 2008, 11:02 AM

Search Tags


/mathhelpforum @mathhelpforum