1. Inequality Help

Im stuck on this.

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?

2. 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.

3. Pick numbers in the intervals $(-\infty, 2), (2,3), (3, \infty)$ 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

4. 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?

5. 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

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