Rationalizing the denominator

[amyers32]

the question, which I am not sure how to type in correctly, is:
25
______
12

all under the square root sign.

So far, I can get : 5
_____
12
with the 12 under the square root sign, but I cannot figure out how to factor out the 12. if you put 4 *3, can't the 4 still be factored out?

I hope this makes sense.

 

[Aladdin]

the question, which I am not sure how to type in correctly, is:
25
______
12

all under the square root sign.

So far, I can get : 5
_____
12
with the 12 under the square root sign, but I cannot figure out how to factor out the 12. if you put 4 *3, can't the 4 still be factored out?

I hope this makes sense.

Your question looks like ?!

\(\displaystyle \frac{5(\sqrt12)}{12}\) ( I've rationalized the denomenator ) -

\(\displaystyle \frac{5(2)\sqrt3}{12}\) Here : \(\displaystyle (\sqrt12=2\sqrt3)\)

\(\displaystyle \frac{5\sqrt3}{6}\)

 

[soroban]

Hello, amyers32!

\(\displaystyle \text{Rationalize: }\;\sqrt{\frac{25}{12}}\)

We are concerned with the denominator; it is not a square.


\(\displaystyle \text{Multiply by }\frac{3}{3}\!:\quad \sqrt{\frac{25}{12}\cdot\frac{3}{3}} \;=\;\sqrt{\frac{75}{36}} \;=\;\frac{\sqrt{75}}{\sqrt{36}} \;=\;\frac{\sqrt{75}}{6}\)


\(\displaystyle \text{Simplify the numerator: }\;\sqrt{75} \;=\;\sqrt{25\cdot3} \;=\;\sqrt{25}\cdot\sqrt{3} \;=\;5\sqrt{3}\)


\(\displaystyle \text{Therefore. we have: }\;\frac{5\sqrt{3}}{6}\)