# Solve the inequality

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• Feb 15th 2008, 04:35 PM
OzzMan
Solve the inequality
Solve the Inequality

$
\left| x^2+x-4\right| <2
$
• Feb 15th 2008, 05:54 PM
mr fantastic
Quote:

Originally Posted by OzzMan
Solve the Inequality

$
\left| x^2+x-4\right| <2
$

I always like to see what's going on with these sorts of equations before I get lost in the algebra.

So here's what I'd do:

Draw a graph of $y = |x^2 + x - 4| = \left| \left( x + \frac{1}{2} \right)^2 - \frac{17}{4} \right|$. It's easy to do .... just draw $y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}$ and reflect in the x-axis the part below the x-axis. Don't be too fussed about where the x-intercepts are, by the way - they are unimportant.

Now draw a graph of $y = 2 = \frac{8}{4}$.

It should be easy now to see that the line y = 2 cuts $y = |x^2 + x - 4|$ at four points. If you know the x-coordinates of those four points you can easily construct the intervals over which $\left| x^2+x-4\right| < 2$ .....

You can see from the graph when you want to solve $x^2+x-4 = 2\,
$
and when you want to solve $-(x^2+x-4) = 2$ in order to get those x-intercepts.
• Feb 15th 2008, 06:00 PM
OzzMan
Quick question. Where exactly did you get $
y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}
$
from?
• Feb 15th 2008, 06:04 PM
mr fantastic
Quote:

Originally Posted by OzzMan
Quick question. Where exactly did you get $
y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}
$
from?

I completed the square on $x^2 + x - 4$:

$x^2 + x - 4 = (x^2 + x) - 4 = \left( x + \frac{1}{2} \right)^2 - \frac{1}{4} - 4 = \left( x + \frac{1}{2} \right)^2 - \frac{1}{4} - \frac{16}{4}$.

Capisce?
• Feb 15th 2008, 06:07 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
[snip]
You can see from the graph when you want to solve $x^2+x-4 = 2\,
$
and when you want to solve $-(x^2+x-4) = 2$ in order to get those x-intercepts.

I'll just add that the part that gets reflected becomes $-(x^2 + x - 4)$. All the other parts are $x^2 + x - 4$.
• Feb 15th 2008, 06:08 PM
OzzMan
I'll show you an example of how I solved this other inequality. Maybe you can tell me if the solving method I used for it is somewhat similar to the first inequality I listed.

$\left| t-27\right| \le 23$

$t-27\le 23\quad \Rightarrow t\le 50$

$-(t-27)\le 23\quad \Rightarrow -t+27\le 23\quad \Rightarrow -t\le -4\quad \Rightarrow t\ge 4$
• Feb 15th 2008, 06:27 PM
mr fantastic
Quote:

Originally Posted by OzzMan
I'll show you an example of how I solved this other inequality. Maybe you can tell me if the solving method I used for it is somewhat similar to the first inequality I listed.

$\left| t-27\right| \le 2$3 Mr F adds ....?

$t-27\le 23\quad \Rightarrow t\le 50$

$-(t-27)\le 23\quad \Rightarrow -t+27\le 23\quad \Rightarrow -t\le -4\quad \Rightarrow t\ge 4$

This is a simple linear inequality. In some ways it essentially boils down to what I've suggested, but I think you're applying the method in a superficial way - without really understanding why it works. That will cause trouble when ......

Things get more complicated with quadratic inequalities. Are you having trouble applying the method I suggested?
• Feb 15th 2008, 06:38 PM
OzzMan
$\left| x^2+x-4\right| <2\quad \Rightarrow x^2+x-6<0\quad and\quad -x^2-x+2<0$

$x^2+x-6=0\quad \Rightarrow (x+3)(x-2)=0\quad \Rightarrow x=-3\quad x=2$

$-x^2-x+2=0\quad \Rightarrow x=\frac{1\pm \sqrt{(1-(4)(-1)(2)}}{-2}\quad \Rightarrow x=\frac{1\pm 3}{-2}\quad \Rightarrow x=-2\quad x=1$

Thats basically the work I have done for this problem. Hows it look?
• Feb 15th 2008, 06:43 PM
mr fantastic
Quote:

Originally Posted by OzzMan
$\left| x^2+x-4\right| <2\quad \Rightarrow x^2+x-6<0\quad and\quad -x^2-x+2<0$

$x^2+x-6=0\quad \Rightarrow (x+3)(x-2)=0\quad \Rightarrow x=-3\quad x=2$

$-x^2-x+2=0\quad \Rightarrow x=1\pm \frac{\sqrt{(1-(4)(-1)(2)}}{-2}\quad \Rightarrow x=\frac{1\pm 3}{-2}\quad \Rightarrow x=-2\quad x=1$

Thats basically the work I have done for this problem.

So it's all blue sky ahead now .... from the graphs that I suggested you draw, do you see that the answer to the question is

-3 < x < -2 or 1 < x < 2.
• Feb 15th 2008, 06:48 PM
OzzMan
Yea I think so. Is there a way to know which way the inequality signs are facing other than graphing? Or is graphing the easiest. I'm all for the easiest method.
• May 13th 2008, 05:18 PM
Krizalid
(I know it's late to post but I leave another way, and of course, it may help someone else.)

Quote:

Originally Posted by OzzMan

Solve the Inequality

$
\left| x^2+x-4\right| <2
$

Inequality equals $-2 Consider the left inequality and rearrange it to $(2x+1)^{2}>9\implies \left| 2x+1 \right|>3.$ From here we have $2x+1>3\implies x>1~\vee~ 2x+1<-3\implies x<-2.$ First solution $S_{1}=(-\infty ,-2)\cup (1,\infty ).$

Let's check the another inequality. (Which is easier to tackle.) $x^{2}+x-4<2\implies (2x+1)^{2}<25\implies \left| 2x+1 \right|<5~\therefore~ -3 Hence, our final solution will be given by (here we only need the $x$ axis for better understanding) $S=(-\infty ,-2)\cup (1,\infty )\cap (-3,\infty )\cap (-\infty ,2)=(-3,-2)\cup (1,2).$