Rationalizing the denominator

amyers32

New member
Joined
Dec 3, 2009
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1
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.
 
amyers32 said:
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 ?!

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

5(2)312\displaystyle \frac{5(2)\sqrt3}{12} Here : (12=23)\displaystyle (\sqrt12=2\sqrt3)

536\displaystyle \frac{5\sqrt3}{6}
 
Hello, amyers32!

Rationalize:   2512\displaystyle \text{Rationalize: }\;\sqrt{\frac{25}{12}}

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


Multiply by 33 ⁣:251233  =  7536  =  7536  =  756\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}


Simplify the numerator:   75  =  253  =  253  =  53\displaystyle \text{Simplify the numerator: }\;\sqrt{75} \;=\;\sqrt{25\cdot3} \;=\;\sqrt{25}\cdot\sqrt{3} \;=\;5\sqrt{3}


Therefore. we have:   536\displaystyle \text{Therefore. we have: }\;\frac{5\sqrt{3}}{6}

 
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