# Solving an inequality

• January 16th 2009, 12:25 AM
Lupin
Solving an inequality
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.
• January 16th 2009, 02:38 AM
Inequalities
Hello Lupin
Quote:

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, 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 and $x>2$, and see what happens to the signs in each case.

I hope you can finish it now.

• January 16th 2009, 05:41 AM
Krizalid
$x^2+x-6<0\implies4x^2+4x-24<0\implies(2x+1)^2<25,$ then $-5<2x+1<5$ so $-3
• January 19th 2009, 08:03 AM
pankaj
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.
• January 19th 2009, 11:20 AM
Soroban
Hello, Lupin!

Quote:

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?

• January 19th 2009, 12:52 PM
Lupin
Thanks for all the help everyone. I totally understand this now. As for Pankaj's questions:

(i) $-5 and $-2

(ii) $-4

I hope they're right, and thanks for the questions.
• January 20th 2009, 03:12 PM
pankaj
Quote:

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

(i) $-5 and $-2

(ii) $-4

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
• February 1st 2009, 08:19 AM
vg284