Solving an inequality

**Lupin** · January 16th 2009, 12:25 AM

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

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

I tried factorising it like this:

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

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.

**Grandad** · January 16th 2009, 02:38 AM

Hello Lupin

Originally Posted by Lupin

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

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

I tried factorising it like this:

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

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 $x$ that make the function zero. Since you've factorised it, that's now very easy: $x = -3$ and $x = 2$ .

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

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

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

I hope you can finish it now.

Grandad

**Krizalid** · January 16th 2009, 05:41 AM

$x^2+x-6<0\implies4x^2+4x-24<0\implies(2x+1)^2<25,$ then $-5<2x+1<5$ so $-3<x<2.$

**pankaj** · January 19th 2009, 08:03 AM

Here is a gift for you Lupin
Solve for x:
(i) $(x-1)(x-2)(x-3)(x+5)<0$

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

I hope you enjoy working on them.

**Soroban** · January 19th 2009, 11:20 AM

Hello, Lupin!

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

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

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

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

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

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?

**Lupin** · January 19th 2009, 12:52 PM

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

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

(ii) $-4<x<7$

I hope they're right, and thanks for the questions.

**pankaj** · January 20th 2009, 03:12 PM

Originally Posted by Lupin

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

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

(ii) $-4<x<7$

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 $x\in(-4,2)\cup(2,7)\cup(11,\infty)$
Remember to exclude 2 since at x=2 LHS becomes 0.Your initial question can be answered in the same manner

**vg284** · February 1st 2009, 08:19 AM

read this doc file

Thread: Solving an inequality

LinkBack

Thread Tools

Display

Solving an inequality

Inequalities

Similar Math Help Forum Discussions

Help solving this inequality

Solving for an inequality

Solving an inequality.

Solving inequality

Solving Inequality

Search Tags