Thread: Radicals - Division

the question is 4/(⁴√64m⁴n²)
the given answer is
(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

Originally Posted by edmundyong

the question is 4/(⁴√64m⁴n²)
the given answer is
(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 ?

yes I dunno how to type the 'root' like that

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.

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

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?

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.

Originally Posted by edmundyong

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

But you can learn to do the mathematical symbols.

[tex]\frac{4}{\sqrt[4]{64m^4n^2}} [/tex] gives
Click on the “go advanced” tab. On the toolbar you will see clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.

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
???