# Finding the set of values for x

• thomas49th
In summary, the set of values for x which satisfies the inequality \frac{(x-3)^{2}}{x+1} < 2 is (-\infty, -1) \cup (1, 7).

## Homework Statement

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??

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.

Hi thomas49th!
thomas49th said:
Find the set of values for x which

(x-3)2/(x+1) < 2

I turned this into

(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?

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)

Sorry I meant values NOT angles. My bad :S

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.
Never done the sets like that. Pretty sure I've never been taught that method, though I can't remeber the method so that's why I am asking :)

(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)

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... I am sure there was a method of drawing out a table?

Thanks :)

thomas49th said:
I don't know why you can multiply them like that. (x+1) is the denominator ! you could say (x+1)-1 maybe?

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).
The way I say it outloud is …

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 … ?

sorted. Thanks :)

## 1. How do I find the set of values for x in an equation?

The set of values for x can be found by solving the equation for x. This can be done by using algebraic operations to isolate x on one side of the equation. Once x is isolated, the resulting expression will represent the set of values for x.

## 2. Can there be more than one set of values for x in an equation?

Yes, there can be more than one set of values for x in an equation. This is especially true for equations with variables such as exponents or radicals, where there could be multiple solutions for x.

## 3. Is it possible for the set of values for x to be infinite?

Yes, in some cases, the set of values for x can be infinite. This can occur when the equation has a variable in the denominator or when there are multiple solutions for x.

## 4. What should I do if I encounter an equation with no solutions for x?

If an equation has no solutions for x, it means that there is no value of x that satisfies the equation. In this case, the solution set for x would be empty, and there is no need to find a set of values for x.

## 5. Can I use a calculator to find the set of values for x?

While a calculator can be useful for solving equations, it is important to understand the steps involved in finding the set of values for x. Relying solely on a calculator may make it difficult to identify any errors or mistakes made in the solution process.