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

Mathematics
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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
you can do this

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Other answers:

4 and 8
\[\sqrt{128x^{5}y^2}\] 128 can be divided as 2*64 or \[2*8^{2}\] and x^5 as \[(x^2)^2*x\]
So\[\sqrt{2*8^2*(x^2)^2*x*y^2}\]
^^^ thats the answer ? im so confused
4 and 8? no see sandeep solved it there
no thats not the answer
you take complete square terms outside like 64 = 8^2 when you take it out of root you get 8
when you take y^2 out of root you get y
and when you take x^4 out of root you get x^2
we have a relation \[\sqrt{a*b*c} = \sqrt{a}*\sqrt{b}*\sqrt{c}\]
Im so confused/:
\[\sqrt{2*8^2*(x^2)^2*x*y^2}\] = \[\sqrt{2}*\sqrt{8^2}*\sqrt{(x^2)^2}*\sqrt{x}*\sqrt{y^2}\]
that gives u \[\sqrt{2}*8*x^2*\sqrt{x}*y\]
\[8\sqrt{2}x^{3/2}y\]
@miah did u get it?
hmm could hv been easier to teach face to face answer is \[8x^2y \sqrt{2xy}\]
Yes i too feel same @aajugdar
Yes it would have been easier. But i some what get it.
smwhat is better than nothing

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