#### madelynnnnnn

##### New member

- Joined
- Apr 6, 2006

- Messages
- 12

ok the question is

sqrt 4/ sqrt 27

then i got

sqrt 108/ 27

then what?

help me!! gracias

sqrt 4/ sqrt 27

then i got

sqrt 108/ 27

then what?

help me!! gracias

- Thread starter madelynnnnnn
- Start date

- Joined
- Apr 6, 2006

- Messages
- 12

ok the question is

sqrt 4/ sqrt 27

then i got

sqrt 108/ 27

then what?

help me!! gracias

sqrt 4/ sqrt 27

then i got

sqrt 108/ 27

then what?

help me!! gracias

What's the square root of 4, madelyn? 2, right?madelynnnnnn said:ok the question is

sqrt 4/ sqrt 27

then i got

sqrt 108/ 27

then what?

help me!! gracias

so sqrt(4) / sqrt(27) = 2 / sqrt(27)

I'll let you finish it...

What you did is absolutely correct!The question is: \(\displaystyle \L\,\frac{\sqrt{4}}{\sqrt{27 }}\)

then i got: \(\displaystyle \L\,\frac{\sqrt{108}}{27}\;\) . . . right!

then what?

Now see if we can simplify \(\displaystyle \,\sqrt{108}\)

\(\displaystyle \;\;\sqrt{108}\;=\;\sqrt{36\cdot3}\;=\;\sqrt{36}\cdot\sqrt{3}\;=\;6\sqrt{3}\)

So we have: \(\displaystyle \L\,\frac{6\sqrt{3}}{27}\;=\;\frac{2\sqrt{3}}{9}\)

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It would be easier (I think) if you simplified first ... as Denis suggested.

You know that: \(\displaystyle \,\sqrt{4}\,=\,2\)

\(\displaystyle \;\;\)and that: \(\displaystyle \,\sqrt{27}\,=\,\sqrt{9\cdot3}\,=\,\sqrt{9}\cdot\sqrt{3}\,=\,3\sqrt{3}\)

So the problem becomes: \(\displaystyle \L\,\frac{2}{3\sqrt{3}}\)

Now rationalize: \(\displaystyle \L\,\frac{2}{3\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\:=\:\frac{2\sqrt{3}}{9}\)

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Always simplify first . . . things usually work out much simpler.

Example: \(\displaystyle \L\,\frac{\sqrt{72}}{\sqrt{50}}\)

Instead of messing around with \(\displaystyle 50\) and \(\displaystyle 3600\),

\(\displaystyle \;\;\sqrt{72}\:=\:\sqrt{36\cdot2}\:=\:\sqrt{6}\sqrt{2}\:=\:6\sqrt{2}\)

\(\displaystyle \;\;\sqrt{50}\:=\:\sqrt{25\cdot2}\:=\:\sqrt{25}\cdot\sqrt{2}\:=\:5\sqrt{2}\)

The problem becomes: \(\displaystyle \L\;\frac{6\sqrt{2}}{5\sqrt{2}}\;=\;\frac{6}{5}\;\;\) . . . see?