# Inequality Help

• Apr 28th 2008, 12:52 PM
nugiboy
Inequality Help
Im stuck on this.

http://i137.photobucket.com/albums/q...tiger/ineq.jpg

I factorised it to (x-3)(x-2)

so there for

x = 3
x = 2

Im not sure how to find the values for which it is smaller/equal to 0.

Any suggestions?
• Apr 28th 2008, 01:01 PM
housefire
Now you know where it crosses the x-axis, if you draw a quick sketch of the graph, you'll be able to see easily where the values you want lie.
• Apr 28th 2008, 01:02 PM
o_O
Pick numbers in the intervals $\displaystyle (-\infty, 2), (2,3), (3, \infty)$ and test them to see if any of them satisfies your inequality. The one that does is representative of all the numbers on that interval.
Code:

   <---------|---------|--------->           2        3
• Apr 28th 2008, 01:05 PM
nugiboy
Quote:

Originally Posted by housefire
Now you know where it crosses the x-axis, if you draw a quick sketch of the graph, you'll be able to see easily where the values you want lie.

Yeh thanks that works. I got it as between 2 and 3, however that does depend on whether the quadratic has its turning point at the top or the bottom. Is there any way to tell from the equation?
• Apr 28th 2008, 01:06 PM
Moo
Hello,

If you have two numbers a and b, the product ab is negative if and only if a and b are of different signs.

This means that if a<0, b>0 and if a>0, b<0 :)
• Apr 28th 2008, 03:40 PM
Krizalid
\displaystyle \begin{aligned} x^{2}-5x+6&\le 0 \\ 4x^{2}-20x+24&\le 0 \\ (2x-5)^{2}&\le 1 \\ \left| 2x-5 \right|&\le 1, \end{aligned}

hence $\displaystyle -1\le 2x-5\le 1\implies 4\le 2x\le 6\,\therefore \,2\le x\le 3.$