How do I simplify this expression: √3 / √3 + √2? The answer is 3 - √6... how?

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How do I simplify this expression: √3 / √3 + √2? The answer is 3 - √6... how?

Mathematics
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is sqrt(2) in the denominator?
no, there's no root 2 in the denominator
\(\sqrt 3/( \sqrt 3+ \sqrt 2)\) now multiply numerator and denominator by \(\sqrt 3- \sqrt 2\) we have \( \sqrt 3/( \sqrt 3 + \sqrt 2) *( \sqrt 3 - \sqrt 2)/( \sqrt 3 - \sqrt 2)\) we get \[ (3-2 \sqrt 6)/( 3-2)\] or \[3-2\sqrt 6\]

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sorry it'd be \[3-\sqrt 6\]
why is it 3 - 2√6?
ignore the 2 int the last two steps
did you get??
but why is it 3 - √6? Did you cancel something?
when we multiplied \(\sqrt 3 -\sqrt 2\) in the numerator and denominator , denominator became 1 . and numerator became \[3-\sqrt 6\]
can you please show me? I'm still confused about what you just said.. so sorry D:
|dw:1327812623013:dw|
It's called multiplying by the conjugate to rationalize the denominator
I have it on my textbook but it's kind of confusing..
because (a-b)(a+b)=a^2-b^2. Thus you eliminate the square roots from the denominator
ohh..
|dw:1327812809044:dw|
the numerator will become = 3- \(\sqrt 6\) denominatorw will be = 3-2= 1
millie did you get ??
ohhh.. kind of, aww man.
why did you square it?
when ever denominator has the form \(\sqrt a\pm \sqrt b\) multiply numerator and denominator by \(\sqrt a\mp \sqrt b\)
ohh, so the opposite.. and then?
we'd in the denominator \[(\sqrt 3 + \sqrt 2)*( \sqrt 3 - \sqrt 2)\] so this is in the form of (a+b)*(a-b)= a^2-b^2 so it'll become \[({\sqrt 3}^2 - {\sqrt 2}^2)\] which is 3-2= 1
ohhh.. it's 3-2= 1? Not 3^2 - 2^2..?
yeah , square of \(\sqrt a\) = a
square root of a number squared is the number itself
milie you got it??
yes I did! thank you!
good :)

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