amonoconnor
  • amonoconnor
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!
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mertsj
  • Mertsj
\[\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
Mertsj
  • Mertsj
Is that the problem?
amonoconnor
  • amonoconnor
Yes!

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Mertsj
  • Mertsj
In the denominator of the first fraction: is it -1or +1
jackthegreatest
  • jackthegreatest
well for one u can cross out the equation on the right, the top and bottom r exactly the same
amonoconnor
  • amonoconnor
Shoot.. you are correct. "-1"
Mertsj
  • Mertsj
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
  • amonoconnor
I know, that's the point: Multiplying by the conjugate, but I'm weary as to how to multiple that across... ?
Mertsj
  • Mertsj
So the denominator is 3-x-1or 2-x
Mertsj
  • Mertsj
now let's do the numerator multiplication:
amonoconnor
  • amonoconnor
How did you get that? :/ I'm not disagreeing, I just don't know how you worked it out..
Mertsj
  • Mertsj
\[(\sqrt{6-x}-2)(\sqrt{3-x}+1)=\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2\]
Mertsj
  • Mertsj
What is (a-b)(a+b)
amonoconnor
  • amonoconnor
(18 - 9x + x) :)
Mertsj
  • Mertsj
Where did that come from?
amonoconnor
  • amonoconnor
Isn't that what'll be under the radical, in the first term in the numerator? :/
Mertsj
  • Mertsj
It would be 18-9x+x^2
amonoconnor
  • amonoconnor
Dude, I'm tired.. my bad again.
Mertsj
  • Mertsj
so put that numerator over 2-x and you are done.
amonoconnor
  • amonoconnor
Can we cancel out anything else?
Mertsj
  • Mertsj
Are there any common factors?
amonoconnor
  • amonoconnor
|dw:1442976045683:dw|
amonoconnor
  • amonoconnor
Is this all correct, and the final version?
Mertsj
  • Mertsj
yep
amonoconnor
  • amonoconnor
|dw:1442976240464:dw|
amonoconnor
  • amonoconnor
This is totally wrong, right?
Mertsj
  • Mertsj
it is correct.
amonoconnor
  • amonoconnor
The second version is correct as well??
amonoconnor
  • amonoconnor
That was my initial, instinctual way to multiple the numerator across, but I thought for sure it was wrong, and I understand how you first showed me, foiling the polynomials UNDER the radical, getting 18 - 9x + x^2. But are you saying that my instinctual way works?

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