Rationalizing Denominator: What Went Wrong in Multiplying by Conjugate?

[temaire]

Homework Statement

Rationalize the denominator and simplify:

$$\frac{x^2-9}{\sqrt{3-x}}$$

Homework Equations

None.

The Attempt at a Solution

The answer to the question is $$-(x+3)\sqrt{3-x}$$, but this is what I'am getting:
http://img222.imageshack.us/img222/9554/radicalio9.jpg
Can someone show me where I went wrong?

 

[rocomath]

Well, when you rationalize a problem. You want to get rid of the radical. In order for the radical to go away, you have to manipulate it so that the power is 1.

You multiplied the both numerator and denominator by $$\sqrt{3+x}$$ when you should have multiplied both by $$\sqrt{3-x}$$

 

[temaire]

But I thought that when you rationalize, you're supposed to multiply the numerator and denominator by the conjugate.

 

[Gib Z]

Since we have a square root, it's like a grouping symbol, we have to take the entire piece as one single thing. So we we let $\sqrt{3+x} = a$, then there's really only one thing in the denominator. We only multiply by the conjugate when the identity $p^2-q^2 = (p+q)(p-q)$ is useful to us, in this case not. This time it was easier just to get rid of the square root.

Or another way to think of it, you have a+0 in the denominator, you have to multiply by its conjugate, a-0 = a.

 

[HallsofIvy]

But I thought that when you rationalize, you're supposed to multiply the numerator and denominator by the conjugate.

Yes, and what is the conjugate here? The conjugate of $a+ b\sqrt{c}$ is $a- b\sqrt{c}$. What are a, b, and c here?