# Inequalities.

• Sep 15th 2008, 04:59 AM
Showcase_22
Inequalities.
I have two inequality questions that I need help solving:

http://i116.photobucket.com/albums/o...sproblem54.jpg

As you can see, i've ended up with all the different inequalities but when I combine then I get the wrong answer.

• Sep 15th 2008, 09:44 AM
Soroban
Hello, Showcase_22!

Here's the first one.
. . Your punchlines are a bit off . . .

Quote:

$x - 4 \;< \;x(x-4) \;<\;5$

The left side: . $x - 4 \:<\:x(x-4)$

We have: . $x - 4 \:<\:x^2-4x \quad\Rightarrow\quad 0 \:< \:x^2-5x + 4 \quad\Rightarrow\quad x^2-5x + 4 \:>\:0$

Factor: . $(x - 1)(x - 4) \:>\:0$

There are two cases: [1] both factors are positive, [2] both factors are negative.

. . $[1]\;\;\begin{Bmatrix}x - 1 &> &0 & \Rightarrow & x & > & 1 \\ x - 4 & > & 0 & \Rightarrow & x & > & 4 \end{Bmatrix} \quad\Rightarrow\quad x \:>\:4$

. . $[2]\;\;\begin{Bmatrix}x - 1 &<&0 & \Rightarrow & x &<& 1 \\ x - 4 &<& 0 & \Rightarrow & x &<& 4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\: 1$

Therefore: . $\boxed{x < 1\;\text{ or }\;x > 4}$

The right side: . $x(x-4) \:<\:5 \quad\Rightarrow\quad x^2 - 4x - 5 \:<\:0$

Factor: . $(x+1)(x-5) \:<\:0$

There are two cases: [1] the factors are (-)(+), [2] the factors are (+)(-).

$[1]\;\;\begin{Bmatrix}x + 1 &<& 0 & \Rightarrow & x &<& \text{-}1 \\
x - 5 &>& 0 & \Rightarrow & x &>& 5 \end{Bmatrix}\quad\Rightarrow\quad \text{impossible}$

$[2]\;\;\begin{Bmatrix}x + 1 &>& 0 & \Rightarrow & x &>& \text{-}1 \\
x-5 &<&0 & \Rightarrow & x &<& 5 \end{Bmatrix} \quad\Rightarrow\quad \boxed{\text{-}1 \:< \:x \:<\:5}$

To determine the final solution, I had to make pretty pictures . . .
Code:

               x < 1  or  x > 4         = = = = o - - - - - o = = = =                 1          4                   -1 < x < 5       - - o = = = = = = = = = = = o - -         -1                      5

Combining them, we have:
Code:

       - - o = = o - - - - - o = = o - -         -1    1          4    5

Therefore: . $\boxed{\;(\text{-}1 \:<\: x\:<\:1)\;\text{ or }\;(4 \:<\:x\:<\:5)\;}$

• Sep 15th 2008, 10:35 AM
Moo
Hi (Happy)

For the second one... ouch !

You have $\frac{|x|-3}{-x^2-1} \ge 1$

Note that $-x^2-1=-(x^2+1)$. Therefore, this is always negative !

And remember, when you multiply an inequality by a negative number, its direction is changed. This means that if you multiply both sides by $-x^2-1$, you'll have :

$|x|-3 ~{\color{red} \le} -(x^2+1)$

Now, see the two situations $|x|=-x$ (if x<0) and $|x|=x$ (if x>0)
If you're not sure, you can do the special case x=0...
• Sep 16th 2008, 04:55 AM
Showcase_22
Thanks Soroban and Moo. I get what I was doing wrong and thanks!

I was comparing my worng answers with the worked solutions I had on another piece of paper. I couldn't follow their method since they were doing some odd stuff with moduli (not that odd now i'm comparing it with correct answers (Giggle)).