Brooklyn

New member
(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

Moderator
Staff member
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

Guest
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

New member
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

Moderator
Staff member
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

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

tkhunny

Moderator
Staff member
"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.