How do you simply the following expression:
(((sqrt(6-x))-2)/((sqrt(3-x)-1)))*(((sqrt(3-x))+1)/((sqrt(3-x))+1)))

Thank you very much! Any and all help is greatly appreciated!

Question

How do you simply the following expression:
(((sqrt(6-x))-2)/((sqrt(3-x)-1)))*(((sqrt(3-x))+1)/((sqrt(3-x))+1)))

Thank you very much! Any and all help is greatly appreciated!

mertsj · Answer

$$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$

mertsj · Answer

Is that the problem?

amonoconnor · Answer

Yes!

mertsj · Answer

In the denominator of the first fraction:  is it -1or +1

jackthegreatest · Answer

well for one u can cross out the equation on the right, the top and bottom r exactly the same

amonoconnor · Answer

Shoot.. you are correct. "-1"

mertsj · Answer

If the problem is as I have posted it the denominators are of the form (a-b)(a+b) and that is equal to a^2-b^2

amonoconnor · Answer

I know, that's the point: Multiplying by the conjugate, but I'm weary as to how to multiple that across... ?

mertsj · Answer

So the denominator is 3-x-1or 2-x

mertsj · Answer

now let's do the numerator multiplication:

amonoconnor · Answer

How did you get that? :/ I'm not disagreeing, I just don't know how you worked it out..

mertsj · Answer

$$(\sqrt{6-x}-2)(\sqrt{3-x}+1)=\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2$$

mertsj · Answer

What is (a-b)(a+b)

amonoconnor · Answer

(18 - 9x + x) :)

mertsj · Answer

Where did that come from?

amonoconnor · Answer

Isn't that what'll be under the radical, in the first term in the numerator? :/

mertsj · Answer

It would be 18-9x+x^2

amonoconnor · Answer

Dude, I'm tired.. my bad again.

mertsj · Answer

so put that numerator over 2-x and you are done.

amonoconnor · Answer

Can we cancel out anything else?

mertsj · Answer

Are there any common factors?

amonoconnor · Answer

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