# Finding the set of values for x

1. Apr 25, 2009

### thomas49th

1. The problem statement, all variables and given/known data
Find the set of values for x which

$$\frac{(x-3)^{2}}{x+1} < 2$$

I turned this into

$$\frac{(x-7)(x-1)}{x+1} < 0$$

Now this implies that the critical angles are x = 7, 1 and -1 but how do I create the sets. I think I have to do somthing with sign change?

Thanks :)

EDIT: WHAT'S WRONG WITH MY LATEX??

2. Apr 25, 2009

### slider142

There is nothing wrong with your latex; the forum is having problems at the moment.

It is a product of 3 terms, so you have the conditions abc < 0 iff (ab < 0 and c > 0) or (bc < 0 and a > 0) or (ac < 0 and b > 0). Inequalities usually generate logical sets like this.

3. Apr 25, 2009

### tiny-tim

Hi thomas49th!
Yes, that's fine …

(and of course it's the same as the product (x-7)(x-1)(x+1) < 0 … do you see why? )

and you know that the sign can only change at 7 1 or -1 (but why are you calling them angles? ), and will change each time (since there are no squared factors)

4. Apr 25, 2009

### thomas49th

Sorry I meant values NOT angles. My bad :S

Never done the sets like that. Pretty sure i've never been taught that method, though I cant remeber the method so that's why im asking :)

I don't know why you can multiply them like that. (x+1) is the denominator !!! you could say (x+1)-1 maybe?

The way I say it outloud is:

y values of (x-7)(x-1) are divide by the y values x+1 graph and I want to find where the new composite graph is below the x-axis. I'm not sure what the most sensible way of doing it is... Im sure there was a method of drawing out a table???

Thanks :)

5. Apr 25, 2009

### tiny-tim

Because (x+1)2 is always positive, so A/(x+1) has the same sign as (A/(x+1))(x+1)2, = A(x+1).
Why are you making this so complicated?

Use slider142's method …

the product of three things will be negative if exactly an odd number of them is negative … ie if either one is negative, or all three are negative …

so in this case … ?

6. Apr 26, 2009

### thomas49th

sorted. Thanks :)