anonymous
  • anonymous
what is the simplest form of √128x^5y^2
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
@aajugdar
anonymous
  • anonymous
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
anonymous
  • anonymous
you can do this

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

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