## anonymous one year ago HELP PLEASE!!!!!! I know the answer I just need to know about the exponents...

1. anonymous

$\sqrt{72x ^{3}y ^{16}}$

2. Nnesha

you can convert square root of something to 1/2 exponents right $\huge\rm \sqrt{x} =x^\frac{ 1 }{ 2 }$ what are the factors of 75 ?

3. Nnesha

x^3 can be written as x^2 times x $\huge\rm \sqrt{72 · x^2 · x· y^{16}} = \sqrt{72x} · x^\frac{ 2 }{ 2 } y^\frac{ 16 }{ 2 }$

4. Nnesha

let me know if you have question about that^^^^

5. anonymous

why do you separate the x^2 and x

6. Nnesha

so i can take square root of x^2 we have to take out the variables from the square root or another way of taking square root of x^3 would |dw:1441409742550:dw|

7. anonymous

and the Y too can you explain the Y why it goes outside

8. Nnesha

how many times does 2 goes into 3 ? one time right and remainder would be one |dw:1441409878485:dw|

9. Nnesha

like i said we can convert root to exponent form $\huge\rm \sqrt[n]{x^m} = x^\frac{ m }{ n }$

10. anonymous

yea

11. anonymous

12. Nnesha

16 is divisible by 2 so i just convert square root to 1/2 exponent form |dw:1441410065097:dw|

13. anonymous

so if it can be divided it's automatically on the outside of the root?

14. phi

the easiest way to do these is as Nnesha showed. change square root to "exponent form" $\sqrt{y^{16} }= \left(y^{16}\right)^\frac{1}{2}$ now use the rule $$(a^b)^c= a^{bc}$$ to write that as $$y^{16\cdot \frac{1}{2}}= y^8$$ notice there is nothing strange (no square roots) it is just y^8

15. anonymous

yea

16. Nnesha

yea here are some example $\huge\rm \sqrt{a^8} = a^\frac{ 8 }{ 2 }=y^{4}$ $\huge\rm \sqrt[3]{y^9} =y^\frac{ 9 }{ 3 }=y^{3}$ of the exponent of the variable not divisible by the root like for x^3 you can still write it as $\huge\rm \sqrt{x^3} =x^\frac{ 3 }{ 2 }$ but the remainder is one so that's mean one x would remain under the root hope it makes sense

17. Nnesha

if*

18. phi

the other way to think of it is $\sqrt{x \cdot x} = x$ if instead of x we had y^8 $\sqrt{y^8 \cdot y^8} = y^8$ that is the same pattern (and of course you know inside the square root y^8 * y^8 is y^16 so $\sqrt{y^{16}}=\sqrt{y^8 \cdot y^8} = y^8$

19. anonymous

okay so i guess i was confused about the 16 cuz i thought it would be a y^4 because 4^2=16 but I think i got it now

20. phi

yes, it can get confusing . the 16 is an exponent. that is different from an ordinary number 16. if you get confused , remember y^16 is short for y*y*y*y..*y (y times itself 16 times) the square root of that is 1/2 of those y's or y*y*y*y*y*y*y*y (y times itself 8 times)

21. anonymous

got it now thanks phi!!