# Absolute Value problem

• Jul 3rd 2010, 11:41 AM
jayshizwiz
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??
• Jul 3rd 2010, 12:48 PM
Unknown008
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.
• Jul 3rd 2010, 01:24 PM
Soroban
Hello, jayshizwiz!

Quote:

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

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

• Jul 3rd 2010, 02:05 PM
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??
• Jul 3rd 2010, 07:09 PM
Prove It
Quote:

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"?
• Jul 4th 2010, 03:29 AM
Unknown008
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.