Question

what is the simplest form of √128x^5y^2

Answer 1

@aajugdar

Answer 2

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

Answer 3

you can do this

Answer 4

4 and 8

Answer 5

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

Answer 6

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

Answer 7

^^^ thats the answer ?
im so confused

Answer 8

4 and 8?
no
see sandeep solved it there

Answer 9

no thats not the answer

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Im so confused/:

Answer 15

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

Answer 16

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

Answer 17

\[8\sqrt{2}x^{3/2}y\]

Answer 18

@miah did u get it?

Answer 19

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

Answer 20

Yes i too feel same @aajugdar

Answer 21

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

Answer 22

smwhat is better than nothing