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Thread: Solving an inequality

  1. #1
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    Solving an inequality

    Can anyone tell me how to solve an inequality like this?

    $\displaystyle x^2+x-6<0$.

    I tried factorising it like this:

    $\displaystyle (x+3)(x-2)<0$

    But I don't see how I can solve it from there. I'd just like to know the method please, so I can do it myself thank you.
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  2. #2
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    Inequalities

    Hello Lupin
    Quote Originally Posted by Lupin View Post
    Can anyone tell me how to solve an inequality like this?

    $\displaystyle x^2+x-6<0$.

    I tried factorising it like this:

    $\displaystyle (x+3)(x-2)<0$

    But I don't see how I can solve it from there. I'd just like to know the method please, so I can do it myself thank you.
    First, you need to find the values of $\displaystyle x$
    that make the function zero. Since you've factorised it, that's now very easy: $\displaystyle x = -3$ and $\displaystyle x = 2$.

    Next, imagine the number line with these numbers
    $\displaystyle -3$ and $\displaystyle 2$ marked, and the number $\displaystyle x$ moving from left to right along the line. When it's right over on the left-hand side, $\displaystyle x < -3$. When it's somewhere in the middle $\displaystyle -3<x<2$, and when it's going off at the right-hand end, $\displaystyle x > 2$. So, look at the signs of your factors $\displaystyle (x+3)$ and $\displaystyle (x-2)$ in each of these three ranges of values of $\displaystyle x$. For instance:

    When $\displaystyle x < -3$ (e.g. $\displaystyle x = -10$), $\displaystyle (x+3)$ is negative and $\displaystyle (x-2)$ is also negative. And a negative times a negative equals a positive. So $\displaystyle (x+3)(x-2)$ is positive. Which is not what we want: we want values of $\displaystyle x$ that make $\displaystyle (x+3)(x-2)<0$.

    Now look at the other two ranges: $\displaystyle -3<x<2$ and $\displaystyle x>2$, and see what happens to the signs in each case.

    I hope you can finish it now.

    Grandad

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  3. #3
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    $\displaystyle x^2+x-6<0\implies4x^2+4x-24<0\implies(2x+1)^2<25,$ then $\displaystyle -5<2x+1<5$ so $\displaystyle -3<x<2.$
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  4. #4
    Senior Member pankaj's Avatar
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    Here is a gift for you Lupin
    Solve for x:
    (i)$\displaystyle (x-1)(x-2)(x-3)(x+5)<0$

    (ii)$\displaystyle (x-2)^2 (x+4)^3(x-7)(x-11)>0$

    I hope you enjoy working on them.
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  5. #5
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    Hello, Lupin!


    Can anyone tell me how to solve an inequality like this?

    $\displaystyle x^2+x-6\:<\:0$.

    I tried factorising it like this: .$\displaystyle (x+3)(x-2)\:<\:0$

    But I don't see how I can solve it from there.

    We have a parabola: .$\displaystyle y \:=\:x^2 + x - 6 \:=\:(x+3)(x-2)$

    It opens upward and has x-intercepts -3 and 2.
    So it looks like this:
    Code:
                    |
          *         |      *
                    |
           *        |    *
        - - * - - - + - * - - - - -
           -3 *     | * 2
                  * |
                    |

    And they are asking: When is the function negative?

    . . That is: When is the parabola below the x-axis?

    Got it?

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  6. #6
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    Thanks for all the help everyone. I totally understand this now. As for Pankaj's questions:

    (i) $\displaystyle -5<x<1$ and $\displaystyle -2<x<3 $

    (ii) $\displaystyle -4<x<7$

    I hope they're right, and thanks for the questions.
    Last edited by Lupin; Jan 19th 2009 at 06:16 PM.
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  7. #7
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Lupin View Post
    Thanks for all the help everyone. I totally understand this now. As for Pankaj's questions:

    (i) $\displaystyle -5<x<1$ and $\displaystyle -2<x<3 $

    (ii) $\displaystyle -4<x<7$

    I hope they're right, and thanks for the questions.
    (i)There appears to be a typo .In place of -2 you should have written 2.

    (ii)There is a very simple way out to do such questions.Plot all the numbers on the number line where the factors become zero.Here we see that -4,2,7,11 are the numbers.
    The number line gets split into 5 parts/intervals.
    (-) .........(+) .........(+)............... (-)................ (+)
    --------------------------------------------------------------Number line
    ........-4 ..........2....... .......7 ...............11...........

    Starting from the rightmost interval start putting the (+) and (-) signs alternatively.But remember sign will not change across that number whose corresponding factor has even power,2 in this case.And you get the desired intervals which are $\displaystyle x\in(-4,2)\cup(2,7)\cup(11,\infty)$
    Remember to exclude 2 since at x=2 LHS becomes 0.Your initial question can be answered in the same manner
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  8. #8
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    read this doc file
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