Math problem, read inside

[Nocturnal]

I'm doing radicals/square roots right now in Algebra.

I have a quick question:

This is what I end up with:

3 Radical 5
/
xy^2

Then I have to rationalize a denominator.

So I end up with:

3sqr(5) · xy^2
/
xy^2 · xy^2

The answer in the book comes out to:

3sqr(5x)
/
xy^2

But I'm trying to figure out where in God's name did the book do away with the y^2.

TIA.

 

[Chaotic42]

You have:

3*sqr(5)
-----------
xy²


The book has:

3*sqr(5x)
------------
xy²

Is that correct?

Edit: What is (x)? Is that multiplication? Using x for multiplication will get you in trouble anywhere outside of middle school.

 

[Nocturnal]

Yes, correct but I have 3*sqr(5xy) due to the rationalization of xy^2. And the book just has (5x) instead of (5xy). I want to know HOW did the y^2 disappear. I know with variables you're supposed to use the formula of a^2 · 1/2 over 1.

 

[Chaotic42]

Why are you multiplying both sides by xy²?

That will give you:

3*sqr(5)*xy²
-----------------
x²y^4

You'd need to multiply both sides by:

1
--
xy²

Which won't do you any good. The only way you're going to get rid of the division is to go with:

3*sqr(5)*(x^-1)*(y^-2)

Edit: Your denominator is already rationalized. An "irrational" denominator would have something like sqr(x) in it. You have no roots in your denominator.

 

[Chaotic42]

Did you ever figure it out?

 

[godspeedx]

You're borked dude.

=)
I don't understand what your problem is.

Ok - I think what you did is simply mess up in the work that you didn't show us. The only difference between what you had originally and what the book gave is the 5 and 5x in the square root. No rationalizing of denominators is needed in that problem.

Try it again and I'll bet you end up with 3 radical 5x / xy^2

 

[DumbGuy]

no roots in denominator, no need to rationalize it

 

[OIKOS]

what math do u have?

 

[Howard]

Can you do this on paper?