• Nov 28th 2012, 06:45 PM
edmundyong
the question is 4/(4√64m4n2)
(4√4n2) / 2mn

but I get 4(4√n2)/ 2mn

4√ = root with 'index=4'

*the index 4 root is for the all
-( 64m4n2)
-(4n2)
-(n2)

I'm not good at writing mathematics symbols sorry. help please SOS
• Nov 28th 2012, 06:48 PM
Prove It
Quote:

Originally Posted by edmundyong
the question is 4/(4√64m4n2)
(4√4n2) / 2mn

but I get 4(4√n2)/ 2mn

4√ = root with 'index=4'

*the index 4 root is for the all
-( 64m4n2)
-(4n2)
-(n2)

I'm not good at writing mathematics symbols sorry. help please SOS

Hang on, is it \displaystyle \displaystyle \begin{align*} \frac{4}{\sqrt[4]{64m^4n^2}} \end{align*}?
• Nov 28th 2012, 07:03 PM
edmundyong
yes :) I dunno how to type the 'root' like that
• Nov 28th 2012, 07:21 PM
coolge
the fourth-root of (64m^4n^2) = 2m fourth-root(2^2 n^2) = 2m sqrt(2n). Then multipy and divide Nr and Dr by sqrt(2n) you will get the answer.
• Dec 1st 2012, 03:56 AM
edmundyong
so you basically what you do is reduce the 4th-root(22n2 ) into sqrt(2n) by diving all by 2?
• Dec 1st 2012, 03:58 AM
edmundyong
Quote:

Originally Posted by coolge
the fourth-root of (64m^4n^2) = 2m fourth-root(2^2 n^2) = 2m sqrt(2n).....

so basically what you do is reduce the 4th-root(22n2 ) into sqrt(2n) by diving all by 2?
• Dec 1st 2012, 04:00 AM
Prove It
Quote:

Originally Posted by edmundyong
so basically what you do is reduce the 4th-root(22n2 ) into sqrt(2n) by diving all by 2?

No, you take everything to the power of 1/4.
• Dec 1st 2012, 04:17 AM
Plato
Quote:

Originally Posted by edmundyong
the question is 4/(4√64m4n2)
(4√4n2) / 2mn
I'm not good at writing mathematics symbols sorry. help please SOS

But you can learn to do the mathematical symbols.

$$\frac{4}{\sqrt[4]{64m^4n^2}}$$ gives $\displaystyle \frac{4}{\sqrt[4]{64m^4n^2}}$
Click on the “go advanced” tab. On the toolbar you will see $\displaystyle \boxed{\Sigma}$ clicking on that give the LaTeX wraps, . The code goes between them.
• Dec 2nd 2012, 05:42 AM
edmundyong
Quote:

Originally Posted by Prove It
No, you take everything to the power of 1/4.

ok
so it'll be 4/2m sqrt(2n)
and mutiply all by sqrt(2n)
1 question
- is sqrt(2n) x sqrt(2n) = 2n?

if yes then...the wouldn't the Nr become
4 [sqrt(2n)] / 2m(2n)
=4 [sqrt(2n) / 4mn
???