anonymous
  • anonymous
How do I simplify this expression: √3 / √3 + √2? The answer is 3 - √6... how?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
is sqrt(2) in the denominator?
anonymous
  • anonymous
no, there's no root 2 in the denominator
ash2326
  • ash2326
\(\sqrt 3/( \sqrt 3+ \sqrt 2)\) 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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More answers

ash2326
  • ash2326
sorry it'd be \[3-\sqrt 6\]
anonymous
  • anonymous
why is it 3 - 2√6?
ash2326
  • ash2326
ignore the 2 int the last two steps
ash2326
  • ash2326
did you get??
anonymous
  • anonymous
but why is it 3 - √6? Did you cancel something?
ash2326
  • ash2326
when we multiplied \(\sqrt 3 -\sqrt 2\) in the numerator and denominator , denominator became 1 . and numerator became \[3-\sqrt 6\]
anonymous
  • anonymous
can you please show me? I'm still confused about what you just said.. so sorry D:
ash2326
  • ash2326
|dw:1327812623013:dw|
anonymous
  • anonymous
It's called multiplying by the conjugate to rationalize the denominator
anonymous
  • anonymous
I have it on my textbook but it's kind of confusing..
anonymous
  • anonymous
because (a-b)(a+b)=a^2-b^2. Thus you eliminate the square roots from the denominator
anonymous
  • anonymous
ohh..
ash2326
  • ash2326
|dw:1327812809044:dw|
ash2326
  • ash2326
the numerator will become = 3- \(\sqrt 6\) denominatorw will be = 3-2= 1
ash2326
  • ash2326
millie did you get ??
anonymous
  • anonymous
ohhh.. kind of, aww man.
anonymous
  • anonymous
why did you square it?
ash2326
  • ash2326
when ever denominator has the form \(\sqrt a\pm \sqrt b\) multiply numerator and denominator by \(\sqrt a\mp \sqrt b\)
anonymous
  • anonymous
ohh, so the opposite.. and then?
ash2326
  • ash2326
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
anonymous
  • anonymous
ohhh.. it's 3-2= 1? Not 3^2 - 2^2..?
ash2326
  • ash2326
yeah , square of \(\sqrt a\) = a
anonymous
  • anonymous
square root of a number squared is the number itself
ash2326
  • ash2326
milie you got it??
anonymous
  • anonymous
yes I did! thank you!
ash2326
  • ash2326
good :)

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