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



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miah
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@aajugdar

aajugdar
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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

aajugdar
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you can do this

miah
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4 and 8

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

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

miah
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^^^ thats the answer ?
im so confused

aajugdar
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4 and 8?
no
see sandeep solved it there

aajugdar
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no thats not the answer

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

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

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

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

miah
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Im so confused/:

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

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

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

SandeepReddy
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@miah did u get it?

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

SandeepReddy
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Yes i too feel same @aajugdar

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

aajugdar
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smwhat is better than nothing