URGENT: adding and subtracting radicals

Brooklyn

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(1/2 times the square root of 6) minus the square root of 3/2

I have got the first part done ( in the paraenthesis) but I don't get the square root of a fraction! What are the steps(if you could show me on an example with out adding the actual(SP?) problem it would be greatly appreaciated.
 

tkhunny

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This?

\(\displaystyle \frac{1}{2}\cdot\sqrt{6}-\sqrt{\frac{3}{2}}\)

Please demonstrate what it is you mean when you say you have completed the one part.
 
J

JeffM

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This may be a double post. If so, I apologize.

Like tkhunny, I am unsure of what your question is.

However, there is nothing mysterious anout the square root of a fraction. Provided a is a non-negative real number, its square root is defined as the non-negative number b such b * b = a. The definition does not care about fractions or whole numbers.

So the square root of (4/9) is (2/3) because (2/3) * (2/3) = (4/9).
It may help to realize that the square root of (c / d) = (the square root of c) / (the square root of d).
 

Brooklyn

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tkhunny said:
\(\displaystyle \frac{1}{2}\cdot\sqrt{6}-\sqrt{\frac{3}{2}}\)
The 1/2 of the square root of 6 I have figured out. The transforming of the second fraction confuses me.
 

tkhunny

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Transform to what? Are you suffering from enforced denomicalitis? This is an itching that crops up whenever a radical is in the denominator. It's a terrible thing.


Hint: \(\displaystyle \sqrt{\frac{3}{2}} = \sqrt{\frac{3\cdot 2}{2\cdot 2}}\)
 

Brooklyn

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Maybe, I don't know. I just know I'm suppose to change the fraction before it can be square rooted or something.
 

tkhunny

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"to square root" is not a verb.

Find the square root.
Calculate the square root.
Determine the square root.

Various ways to say it.

You must utilize this principle, a/a = 1 for a <> 0 until you get those radical s out fo the denominator. Personally, I do not recommend such practices, but if you have to pass a test with it, then learn it.
 
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