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?

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- Apr 28th 2008, 12:52 PMnugiboyInequality 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 PMhousefire
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 PMo_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 PMnugiboy
- Apr 28th 2008, 01:06 PMMoo
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 PMKrizalid
$\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 $\displaystyle -1\le 2x-5\le 1\implies 4\le 2x\le 6\,\therefore \,2\le x\le 3.$