Solve the Inequality

$\displaystyle

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

$

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- Feb 15th 2008, 04:35 PMOzzManSolve the inequality
Solve the Inequality

$\displaystyle

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

$ - Feb 15th 2008, 05:54 PMmr fantastic
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. - Feb 15th 2008, 06:00 PMOzzMan
Quick question. Where exactly did you get $\displaystyle

y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}

$ from? - Feb 15th 2008, 06:04 PMmr fantastic
- Feb 15th 2008, 06:07 PMmr fantastic
- Feb 15th 2008, 06:08 PMOzzMan
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.

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

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

$\displaystyle -(t-27)\le 23\quad \Rightarrow -t+27\le 23\quad \Rightarrow -t\le -4\quad \Rightarrow t\ge 4$ - Feb 15th 2008, 06:27 PMmr fantastic
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 PMOzzMan
$\displaystyle \left| x^2+x-4\right| <2\quad \Rightarrow x^2+x-6<0\quad and\quad -x^2-x+2<0$

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

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

Thats basically the work I have done for this problem. Hows it look? - Feb 15th 2008, 06:43 PMmr fantastic
- Feb 15th 2008, 06:48 PMOzzMan
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 PMKrizalid
(I know it's late to post but I leave another way, and of course, it may help someone else.)

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) $\displaystyle S=(-\infty ,-2)\cup (1,\infty )\cap (-3,\infty )\cap (-\infty ,2)=(-3,-2)\cup (1,2).$