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Math Help - Solve the inequality

  1. #1
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    Solve the inequality

    Solve the Inequality

    <br />
\left| x^2+x-4\right| <2<br />
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  2. #2
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    Quote Originally Posted by OzzMan View Post
    Solve the Inequality

    <br />
\left| x^2+x-4\right| <2<br />
    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\, <br />
and when you want to solve -(x^2+x-4) = 2 in order to get those x-intercepts.
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  3. #3
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    Quick question. Where exactly did you get <br />
y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}<br />
from?
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  4. #4
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    Quote Originally Posted by OzzMan View Post
    Quick question. Where exactly did you get <br />
y = \left( x + \frac{1}{2} \right)^2 - \frac{17}{4}<br />
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?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    [snip]
    You can see from the graph when you want to solve x^2+x-4 = 2\, <br />
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.
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  6. #6
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    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
    Last edited by OzzMan; February 15th 2008 at 06:39 PM.
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  7. #7
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    Quote Originally Posted by OzzMan View Post
    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 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?
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  8. #8
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    \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?
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  9. #9
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    Quote Originally Posted by OzzMan View Post
    \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.
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  10. #10
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    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.
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  11. #11
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    (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 View Post

    Solve the Inequality

    <br />
\left| x^2+x-4\right| <2<br />
    Inequality equals -2<x^{2}+x-4<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<x<2. 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).
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