1. ## Solve the inequality

Solve the Inequality

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

2. Originally Posted by OzzMan
Solve the Inequality

$\displaystyle \left| x^2+x-4\right| <2$
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 $\displaystyle y = |x^2 + x - 4| = \left| \left( x + \frac{1}{2} \right)^2 - \frac{17}{4} \right|$. It's easy to do .... just draw $\displaystyle y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}$ 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 $\displaystyle y = 2 = \frac{8}{4}$.

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

You can see from the graph when you want to solve $\displaystyle x^2+x-4 = 2\,$ and when you want to solve $\displaystyle -(x^2+x-4) = 2$ in order to get those x-intercepts.

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

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

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

Capisce?

5. Originally Posted by mr fantastic
[snip]
You can see from the graph when you want to solve $\displaystyle x^2+x-4 = 2\,$ and when you want to solve $\displaystyle -(x^2+x-4) = 2$ in order to get those x-intercepts.
I'll just add that the part that gets reflected becomes $\displaystyle -(x^2 + x - 4)$. All the other parts are $\displaystyle x^2 + x - 4$.

6. 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.

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

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

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

7. 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.

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

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

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

8. $\displaystyle \left| x^2+x-4\right| <2\quad \Rightarrow x^2+x-6<0\quad and\quad -x^2-x+2<0$

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

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

Thats basically the work I have done for this problem. Hows it look?

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

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

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

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.

10. 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.

11. (I know it's late to post but I leave another way, and of course, it may help someone else.)

Originally Posted by OzzMan

Solve the Inequality

$\displaystyle \left| x^2+x-4\right| <2$
Inequality equals $\displaystyle -2<x^{2}+x-4<2.$ Consider the left inequality and rearrange it to $\displaystyle (2x+1)^{2}>9\implies \left| 2x+1 \right|>3.$ From here we have $\displaystyle 2x+1>3\implies x>1~\vee~ 2x+1<-3\implies x<-2.$ First solution $\displaystyle S_{1}=(-\infty ,-2)\cup (1,\infty ).$

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