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Math Help - Absolute Value problem

  1. #1
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    Absolute Value problem

    |x^2 - 3x + 2| < |x+2|

    First of all, does anyone know where I can find an explanation of these types of questions on the internet? I've never solved absolute value questions before so I don't know exactly what to do. And my book just solves the question but doesn't explain it at all...

    I understand this much:

    |x+ 2| = \left\{\begin{array}{cc}{x+2}, &\mbox{ if }<br />
x\geq -2\\{-(x+2), & \mbox{ if } x<-2\end{array}\right

    |x^2 - 3x + 2| = |(x-1)(x-2)| = \left\{\begin{array}{cc}{x^2 - 3x + 2},&\mbox{ if }x\geq 2 \mbox{ or }x \leq 1\\{-(x^2 -3x +2)}, & \mbox{ if }  1<x<2\end{array}\right

    And from here I don't understand - the book starts answering the question by the boundaries:

    (i)
    for \mbox{-2} \leq x \leq 1 \mbox{ or } x \geq 2

    the inequality x^2-3x+2 < x+2 \rightarrow x(x-4) < 0 and this inequality produces 0<x<4

    So I won't continue the rest of the answer. If I understand this part I assume I'll understand the rest. How did they get (i) -2 \leq x \leq 1 \mbox{ or } x \geq 2 as the boundary? Where did that inequality come from??
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Well, that's how I do it:

    |x^2 - 3x + 2|<|x+2|

    (x^2 -3x + 2)^2 < (x+2)^2

    Expanding and simplifying,

    x(x-4)(x^2 -2x +4)<0

    Now, since x^2 - 2x + 4 < 0 has no real solutions,

    x(x-4) < 0

    The critical values are x = 0 and x = 4.

    I do my table, and get as answer: 0 < x < 4.
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  3. #3
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    Hello, jayshizwiz!

    How about a graphic solution?


    |x^2 - 3x + 2| \:<\: |x+2|

    The graph of y \:=\:x^2-3x+2 \:=\:(x-1)(x-2)
    . . is an up-opening parabola with x-intercepts 1 and 2.

    Its graph looks like this:

    Code:
            |
           *|                      *
            |
            |
            *                     *
            | 
            |*                   *
            | *                 *
        - - + - * - - - - - - * - - - -
            |   1  *       *  2
            |          *
            |


    The graph of y \:=\:|x^2-3x+2| looks like this:

    Code:
            |
           o|                      o
            |
    
            |
            o                     o
            | 
            |o         o         o
            | o   o         o   o
      - - - + - o - - - - - - o - - - -
            |   1             2
            |
    Everything below the x-axis is reflected upward.



    The graph of y \:=\:x+2 looks like this:
    Code:
                |
                |         *
                |       *
                |     *
                |   *
               2| *
                *
              * |
            *   |
          *     |
    - - * - - - + - - - - - -
      * -2      |
    *           |
                |


    The graph of y \:=\:|x+2| looks like this:

    Code:
                |
                |         ∆
                |       ∆
                |     ∆
                |   ∆
                | ∆
                ∆
              ∆ |
    ∆       ∆   |
      ∆   ∆     |
    - - ∆ - - - + - - - - - -
        -2      |


    Sketch the two graphs on a set of coordinate axes.

    Code:
                |
             o  |                         ♥(4,6)
                |                       ∆
                |                     ∆
                |                   ∆
                |                 ∆
                |               ∆
              o |             ∆         o
                |           ∆
                |         ∆
                |       ∆
               o|     ∆                o
                |   ∆
           (0,2)| ∆
                ♥                     o
              ∆ | 
    ∆       ∆   |o         o         o
      ∆   ∆     | o   o         o   o
    - - ∆ . . - + - o - - - - - - o - - - -
       -2       |   1             2
                |


    When is the "parabola" below the "line"?

    Answer: . 0 \:<\:x\:<\:4

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  4. #4
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    Post

    Thanks guys!

    wow, Soroban your graphs are awesome. And I like how you put little hearts at (4, 6) and (0,2)!

    Just to make sure, is 0<x<4 the only answer you guys got??
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  5. #5
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    Quote Originally Posted by jayshizwiz View Post
    Thanks guys!

    wow, Soroban your graphs are awesome. And I like how you put little hearts at (4, 6) and (0,2)!

    Just to make sure, is 0<x<4 the only answer you guys got??
    Well, is there anywhere else where the "parabola" is less than the "line"?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Well, you can see the graph now. The graph of |x+2| is above that of |x^2 - 3x + 2| only within that range.

    When you go towards the left, the parabolic part of |x^2 - 3x + 2| goes up 'faster' than the graph of |x+2|, hence, |x+2| will never be above the parabola again.

    The same thing occurs on the right.
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