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 Post subject: GRE Algebra
PostPosted: Thu Nov 15, 2012 5:43 pm 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
Please explain how this equation works out:

s² / (s²√3 / 4) = 4s² / (s²√3) = 4 / √3 = (4√3) / 3

I am with each step, crossing out the s², and then you have 4 / √3. But how does 4 / √3 turn into 4√3 / 3? Isn't that an extra three?

I think what I really need is a lesson in multiplying/dividing by radicals. I always seem to mess them up and the only thing I ever read is telling me to group radicals together and treat regular numbers separately. I do not see anything that goes into depth. Please help!


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 Post subject: Re: GRE Algebra
PostPosted: Thu Nov 15, 2012 6:03 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
But how does 4 / √3 turn into 4√3 / 3? Isn't that an extra three?
This is the method to eliminate radicals in a denominator.
We multiply the numerator and denominator by the same number, √3.

4 / √3 = 4√3 / (√3 × √3)

The denominator, √3 × √3, results in (√3)² = 3 . So we get:

4 / √3 = (4√3) / 3

Quote:
I always seem to mess them up and the only thing I ever read is telling me to group radicals together and treat regular numbers separately.
If you are Ok with exponents, then you can treat radicals as exponents.
a is a to the (1/2)th power

Another two simple ideas that should get you started on roots are:
1. √a × √a = a
³√a × ³√a × ³√a = a

2. Roots are Ok to multiply in & out (as long as all the numbers under the sign are non-negative):
√(2/3) = √2 / √3
√6 = √(2 × 3) = √2 × √3

√6 / √2 = √(6/2) = √3
or
√6 / √2 = √3 × √2 / √2 = √3

If you get the feeling of these two basic concepts, that should allow you to move on to dealing with addition. For example we can transform: √6 + √2 = √2 (√3 + 1).


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