Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - Roots of Absolute Value with inequalities

  1. #1
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2

    Roots of Absolute Value with inequalities

    I just wonder why when i solved

    |x+2|^2 < |x( x - 1)|^2
    .
    .
    .( x +2 )^2 - ( x^2 - x) ^2 < 0
    .( 2 + 2x - x^2) ( x^2+2) < 0
    .- ( x^2 - 2x - 2) ( x^2 + 2) < 0
    since it is < 0 then i have to reject the ( x^2 + 2)
    . then i have left (x^2 - 2x - 2) > 0
    . square root of (x - 1)^2 > square root of 3
    . |x -1| > sq. root 3

    then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3

    and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
    that is my problem...
    i need guide here.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Roots of Absolute Value with inequalities

    Quote Originally Posted by rcs View Post
    I just wonder why when i solved

    |x+2|^2 < |x( x - 1)|^2
    .
    .
    .( x +2 )^2 - ( x^2 - x) ^2 < 0
    .( 2 + 2x - x^2) ( x^2+2) < 0
    .- ( x^2 - 2x - 2) ( x^2 + 2) < 0
    since it is < 0 then i have to reject the ( x^2 + 2)
    . then i have left (x^2 - 2x - 2) > 0
    . square root of (x - 1)^2 > square root of 3
    . |x -1| > sq. root 3

    then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3

    and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
    that is my problem...
    i need guide here.

    Thanks
    Once you get to \displaystyle \begin{align*} -\left(x^2 - 2x - 2\right)\left(x^2 + 2\right) &< 0 \end{align*} we have

    \displaystyle \begin{align*} \left(x^2 - 2x - 2\right)\left(x^2 + 2\right) &> 0 \textrm{ and since the second term is always positive, we can divide both sides by it and keep the inequality intact...} \\ x^2 - 2x - 2 &> 0 \\ x^2 - 2x + (-1)^2 - (-1)^2 - 2 &> 0 \\ (x - 1)^2 - 3 &> 0 \\ (x - 1)^2 &> 3 \\ |x - 1| &> \sqrt{3} \\ x - 1 < -\sqrt{3} \textrm{ or } x - 1 &> \sqrt{3} \\ x < 1 - \sqrt{3} \textrm{ or } x &> 1 + \sqrt{3}  \end{align*}
    Thanks from rcs
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466

    Re: Roots of Absolute Value with inequalities

    Quote Originally Posted by rcs View Post
    I just wonder why when i solved

    |x+2|^2 < |x( x - 1)|^2
    .
    .
    .( x +2 )^2 - ( x^2 - x) ^2 < 0
    .( 2 + 2x - x^2) ( x^2+2) < 0
    .- ( x^2 - 2x - 2) ( x^2 + 2) < 0
    since it is < 0 then i have to reject the ( x^2 + 2)
    . then i have left (x^2 - 2x - 2) > 0
    . square root of (x - 1)^2 > square root of 3
    . |x -1| > sq. root 3

    then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3
    If these were correct, you wouldn't need both. "x> 1- sq. root of 3" would be sufficient.

    and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
    that is my problem...
    i need guide here.

    Thanks
    Did you check your answer? If x= 0, it satisfies x< 1+ sqrt(3) so should be in the solution set but |0+2|^2= 4 while |0(0-1)|^2= 0. That does NOT satisfy the inequality. Your error is in thinking that [b]nonlinear[b] inequalities can be treated like linear inequalities. If x^2> a it is NOT true that x>\sqrt{a}.

    The best way to solve more complicated inequalities is to first solve the associated equation. If you solve |x+ 2|^2= |x(x-1)|^2 you get, of course, x= 1+ sqrt(3) and x= 1- sqrt(3). Because the function is continuous, in order to go from "+" to "-", the function value must pass through "0" so those two points separate ">" from "<". Check a single point in each of the three intervals, say, -3< 1- sqrt(3), 1-sqrt(3)< 0< 1+ sqrt(3), and 1+sqrt(3)< 3, to see which satisfy the inequality and which do not.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Absolute value Inequalities 2
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 17th 2011, 02:24 PM
  2. Absolute value inequalities.
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: February 2nd 2011, 04:34 PM
  3. absolute value inequalities
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 1st 2011, 08:49 PM
  4. Absolute Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 29th 2008, 04:35 PM
  5. Absolute Inequalities
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 12th 2007, 10:32 AM

Search Tags


/mathhelpforum @mathhelpforum