# Math Help - Absolute Value problem

1. ## 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:

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

$|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

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

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??

2. 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.

3. Hello, jayshizwiz!

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

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: . $0 \:<\:x\:<\:4$

4. 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??

5. Originally Posted by jayshizwiz
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"?

6. 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.