Results 1 to 5 of 5

Math Help - Inequality

  1. #1
    Member RedBarchetta's Avatar
    Joined
    Apr 2008
    From
    United States
    Posts
    114

    Inequality

    Can anyone tell me how my book went from point a to b?

    <br />
|x^2  - 1| < 2 \Rightarrow  - \sqrt 3  < x < \sqrt 3 <br />

    I just don't see it...thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by RedBarchetta View Post
    Can anyone tell me how my book went from point a to b?

    <br />
|x^2 - 1| < 2 \Rightarrow - \sqrt 3 < x < \sqrt 3 <br />

    I just don't see it...thanks.
    Try drawing the graphs of y = |x^2 - 1| and y = 2. Now calculate the x-coordinate of the intersection points of these two graphs:

    x^2 - 1 = 2 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3} or x = -\sqrt{3}.

    For what values of x is the graph of y = |x^2 - 1| below the graph of y = 2 ....?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member RedBarchetta's Avatar
    Joined
    Apr 2008
    From
    United States
    Posts
    114
    Ah ha, now that makes much more sense graphically.

    How would you solve this without looking at it graphically?

    Using the typical method for solving an inequality doesn't work here.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Why doesn't it work here ?


    |x^2-1|<2 \Leftrightarrow -2<x^2-1<2 \Leftrightarrow -1<x^2<3
    Whatever x is, the latter inequality can be restricted to :
    x^2<3, since x^2 \geq 0.

    So we have this new inequality :
    x^2<3

    which gives |x|< \sqrt{3} (*)

    that is to say -\sqrt{3}<x<\sqrt{3}



    ~~~~~~~~~~~~~~~~~~~~~~
    note : |a|<b \Leftrightarrow -b<a<b, where b is positive.

    (*) this can be proved.
    a>0, x^2<a^2 \Leftrightarrow |x|<a

    x^2-a^2<0
    (x-a)(x+a)<0

    so either x-a<0 and x+a>0 ----> x < a and x > -a. This is equivalent to |x|< a
    either x-a>0 and x+a<0 -----> x > a and x < -a. This is not possible since a > 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member RedBarchetta's Avatar
    Joined
    Apr 2008
    From
    United States
    Posts
    114
    Thanks guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 09:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 02:16 PM
  3. inequality
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: July 25th 2010, 07:11 PM
  4. Inequality help
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 8th 2010, 07:24 AM
  5. Inequality :\
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 12th 2009, 02:57 PM

Search Tags


/mathhelpforum @mathhelpforum