Quotient theorem for square roots

Richay

New member
Joined
Mar 31, 2006
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43
I got 3 out of the 4 problems on this right. Yet i'm still stuck on one.

+


What would be the answer?
 

soroban

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Jan 28, 2005
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5,588
Hello, Richay!

\(\displaystyle \L -3\sqrt{\frac{2}{3}}\,+\,2\sqrt{\frac{3}{2}}\)


Simplify each term:

\(\displaystyle \L\;\;-3\cdot\sqrt{\frac{2}{3}\cdot\frac{3}{3}}\:=\:-3\cdot\sqrt{\frac{6}{9}}\:=\:-3\cdot\frac{\sqrt{6}}{\sqrt{9}} \:=\;-\not{3}\cdot\frac{\sqrt{6}}{\not{3}}\:=\:-\sqrt{6}\)

\(\displaystyle \L\;\;2\cdot\sqrt{\frac{3}{2}\cdot\frac{2}{2}}\:=\:2\cdot\sqrt{\frac{6}{4}}\:=\:2\cdot\frac{\sqrt{6}}{\sqrt{4}}\:=\:\not{2}\cdot\frac{\sqrt{6}}{\not{2}}\:=\;\sqrt{6}\)

The problem becomes: \(\displaystyle \L\,-\sqrt{6}\,+\,\sqrt{6}\:=\:0\)
 

Richay

New member
Joined
Mar 31, 2006
Messages
43
Hey.
Thanks for the explenation.
K i see how it's done :D
 
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