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

[Gliese123]

Homework Statement

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)

The Attempt at a Solution

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

($\sqrt{3-x}$$\sqrt{x+9}$)²=(8x+1)²

Which should lead to this:

(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.

 

[Saitama]

Homework Statement

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)

The Attempt at a Solution

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

($\sqrt{3-x}$$\sqrt{x+9}$)²=(8x+1)²

Which should lead to this:

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

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?

 

[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.

 

[Gliese123]

Thank you for answering.
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?

 

[Gliese123]

-6 & 0 perhaps?

 

[Saitama]

Thank you for answering.
The square of $\sqrt{x}$ is x.

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

Anyways, you need not worry about it at the moment.

What do you mean by the radicals?

By radicals, I mean those square roots and the stuff inside them. For what values of x is $\sqrt{3-x}$ defined?

 

[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?

 

[Saitama]

3? Sorry, I'm retarded. By defined, is it meant to become 0?

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?

 

[Gliese123]

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

 

[Gliese123]

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

 

[Gliese123]

Shouldn't it just be 3 and -9 ?

 

[Saitama]

Shouldn't it just be 3 and -9 ?

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?

 

[BruceW]

Shouldn't it just be 3 and -9 ?

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