# Find the interval for a valid x between a & b.

1. Aug 11, 2013

### Gliese123

1. The problem statement, all variables and given/known data
The left side in the equation $\sqrt{3-x}$$\sqrt{x+9}$=8x+1 is only definable when x is located in a certain interval a $\leq$ x $\leq$ b. What's a & b? (Note: The answer ought to be integers)

3. The attempt at a solution
$\sqrt{3-x}$$\sqrt{x+9}$=8x+1

($\sqrt{3-x}$$\sqrt{x+9}$)2=(8x+1)2

(3-x)*2($\sqrt{3-x}$$\sqrt{x+9}$)(x+9)=64x2+22x-26

The problem now is, what shall I do next? Expand the parentheses?
Well.. If I do that then what? I'll obviously get a quadratic equation. Any advise what to do with the left side? Those roots are hopeless.

2. Aug 11, 2013

### Saitama

No. What is the square of $\sqrt{x}$?

Instead of squaring, look at the radicals first. For what values of x, is $\sqrt{3-x}$ is defined? When $\sqrt{x+9}$ is defined?

3. Aug 11, 2013

### LCKurtz

Also, the way the problem is stated, you aren't asked to solve anything. Just find the interval. And it won't be just integers, unless you just mean the end points are integers.

Last edited: Aug 11, 2013
4. Aug 11, 2013

### Gliese123

The square of $\sqrt{x}$ is x.
What do you mean by the radicals? Like when the two x have a certain value and summarizes to then have the same number under the root?

5. Aug 11, 2013

### Gliese123

-6 & 0 perhaps?

6. Aug 11, 2013

### Saitama

So what is the sqaure of $\sqrt{3-x}\sqrt{x+9}$?

Anyways, you need not worry about it at the moment.
By radicals, I mean those square roots and the stuff inside them. For what values of x is $\sqrt{3-x}$ defined?

7. Aug 11, 2013

### Gliese123

(3-x)(x+9)

By radicals, I mean those square roots and the stuff inside them. For what values of x is √(3-x) defined?
3? Sorry, I'm retarded. By defined, is it meant to become 0?

8. Aug 11, 2013

### Saitama

Nooo.

$\sqrt{3-x}$ is defined for 3 but the question asks for the range of values of x. For $\sqrt{x}$ to be defined, x should either greater than or equal to zero. Do you get an idea now?

9. Aug 11, 2013

### Gliese123

Ahaaaaaa... :uhh: :thumbs:
So 0 ? I don't f**king know. :yuck:

10. Aug 11, 2013

### Gliese123

Sorry. I'll go to sleep. This is too difficult for me. Thank you for the help.

11. Aug 14, 2013

### Gliese123

Shouldn't it just be 3 and -9 ?

12. Aug 14, 2013

### Saitama

What should be "just 3 and -9"? Are you talking about the values of x? Then no. I already said that you need to find the range of values of x.

$\sqrt{3-x}$ is defined when $3-x \geq 0$. Do you understand why?

13. Aug 14, 2013

### BruceW

well, that is correct if you mean they are the endpoints of the 'allowed' interval of x values.