1. Apr 16, 2010

### coverband

1. The problem statement, all variables and given/known data
Find all values of x for which x^2-5x+6<0

2. Relevant equations
(x-2)(x-3)<0

3. The attempt at a solution
When I draw the graph the solution is clearly 2<x<3

However, if I approach it mathematically (x-2)(x-3)<0
This implies either 1. (x-2) is positive and (x-3) is negative OR
2. (x-2) is negative and (x-3) is positive
Case 1 implies 2<x<3
Case 2 implies 2>x>3

Whats up !!!??

2. Apr 16, 2010

### Mentallic

Look at case 2 again. How can x-2 be negative while x-3 is positive? Is there any value of x where this is true?

3. Apr 16, 2010

### irycio

So what's wrong? the correct answer is $$x \in (2,3)$$.
This is what you've got from the picture (probably) and what you've got from your alternative. You yourself have written (...) OR (...). Since case 1 gives you the correct answer, and case 2 is $$\emptyset$$, it's fine.

4. Apr 16, 2010

### coverband

True. Whats the general way of solving these?

5. Apr 16, 2010

### coverband

Can we not solve these exclusively from numbers on a page without graphs

6. Apr 16, 2010

### Mentallic

You need to take 3 cases.

1) When x-2 and x-3 are both positive
2) When x-2 is positive while x-3 is negative (so x is between 2 and 3)
3) When they're both negative.=

7. Apr 16, 2010

### coverband

I don't think so

1) if x-2 and x-3 positive, their product won't give a negative number
2) fair enough but what if we had (x-2)(x+3)<0
3) give a positive?

8. Apr 16, 2010

### irycio

I believe that that was meant to be genral solution.

if you had (x-2)(x+3)<0 then you would just have to take the alternative, as you've done in your firs post. Of course, the result would be different.

9. Apr 16, 2010

### Staff: Mentor

If we're talking about the problem in this thread, (x - 2)(x - 3) < 0, there are only two cases, the ones listed by the OP at the beginning of this thread.

If we're talking more generally, with (x - a)(x - b) > 0 or (x - a)(x - b) < 0, there are four cases.

1) When x-a and x-b are both positive.
2) When x-a is positive and x-b is negative.
3) When x -a is negative and x - b is positive.
3) When x-a and x-b are both negative.

To get back to the original question, where x - 2 and x -3 have to be opposite in sign, we have
1) x -2 is positive and x-3 is negative.
or
2) x -2 is negative and x-3 is positive.

For 1, x - 2 > 0 and x - 3 < 0 <==> x > 2 and x < 3 <==> 2 < x < 3
For 2, x - 2 < 0 and x - 3 > 0 <==> x < 2 and x > 3 This set is empty.

Therefore, x^2 -5x + 6 < 0 <==> 2 < x < 3

10. Apr 16, 2010

### coverband

Thanks. What do you mean "The set is empty"?

11. Apr 16, 2010

### irycio

There are no real (or any other) numbers that would be in tthe same time larger than 3 and lesser than 2. Hance, the set that contains them is empty, as there are no elements in it.

12. Apr 16, 2010

### coverband

Does 2>x>3 not mean x bigger than 3 OR less than 2?how do you know it means AND?

13. Apr 16, 2010

### tiny-tim

Hi coverband!

2=x=y means AND, doesn't it?

So 2>x>y means AND …

generally, a statement like that is a single statement, and it must be entirely true.

14. Apr 16, 2010

### coverband

15. Apr 16, 2010

### malicx

I'm not sure who invented it, but as long as the transitive property holds (which it does for inequalities on real numbers) it seems like a natural extension. It can be proved by the properties of real numbers.

Assume that a < b and that b < c. Add the two inequalities to get

a + b < b + c
a < c

Adding two inequalities together in this way is allowed, but to show that requires another proof!

Last edited: Apr 16, 2010
16. Apr 16, 2010

### Staff: Mentor

No. It means three things:
1. 2 > x
2. x > 3
3. 2 > 3 (due to transitivity of the > relation)
All of these have to be true, which means that the first inequality has to be true AND the second inequality has to be true AND the third inequality has to be true.

When you write 2 > x > 3 you are tacitly saying that 2 > 3, which is of course not true.