## miah 2 years ago what is the simplest form of √128x^5y^2

1. miah

@aajugdar

2. aajugdar

split 128 in such a way that you can get a complete square like in last problem 9 was square of 3 thats why 63=9*7

3. aajugdar

you can do this

4. miah

4 and 8

5. SandeepReddy

$\sqrt{128x^{5}y^2}$ 128 can be divided as 2*64 or $2*8^{2}$ and x^5 as $(x^2)^2*x$

6. SandeepReddy

So$\sqrt{2*8^2*(x^2)^2*x*y^2}$

7. miah

^^^ thats the answer ? im so confused

8. aajugdar

4 and 8? no see sandeep solved it there

9. aajugdar

10. aajugdar

you take complete square terms outside like 64 = 8^2 when you take it out of root you get 8

11. aajugdar

when you take y^2 out of root you get y

12. aajugdar

and when you take x^4 out of root you get x^2

13. SandeepReddy

we have a relation $\sqrt{a*b*c} = \sqrt{a}*\sqrt{b}*\sqrt{c}$

14. miah

Im so confused/:

15. SandeepReddy

$\sqrt{2*8^2*(x^2)^2*x*y^2}$ = $\sqrt{2}*\sqrt{8^2}*\sqrt{(x^2)^2}*\sqrt{x}*\sqrt{y^2}$

16. SandeepReddy

that gives u $\sqrt{2}*8*x^2*\sqrt{x}*y$

17. SandeepReddy

$8\sqrt{2}x^{3/2}y$

18. SandeepReddy

@miah did u get it?

19. aajugdar

hmm could hv been easier to teach face to face answer is $8x^2y \sqrt{2xy}$

20. SandeepReddy

Yes i too feel same @aajugdar

21. miah

Yes it would have been easier. But i some what get it.

22. aajugdar

smwhat is better than nothing