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miah

  • 2 years ago

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

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  1. miah
    • 2 years ago
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    @aajugdar

  2. aajugdar
    • 2 years ago
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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

  3. aajugdar
    • 2 years ago
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    you can do this

  4. miah
    • 2 years ago
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    4 and 8

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

  6. SandeepReddy
    • 2 years ago
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    So\[\sqrt{2*8^2*(x^2)^2*x*y^2}\]

  7. miah
    • 2 years ago
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    ^^^ thats the answer ? im so confused

  8. aajugdar
    • 2 years ago
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    4 and 8? no see sandeep solved it there

  9. aajugdar
    • 2 years ago
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    no thats not the answer

  10. aajugdar
    • 2 years ago
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    you take complete square terms outside like 64 = 8^2 when you take it out of root you get 8

  11. aajugdar
    • 2 years ago
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    when you take y^2 out of root you get y

  12. aajugdar
    • 2 years ago
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    and when you take x^4 out of root you get x^2

  13. SandeepReddy
    • 2 years ago
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    we have a relation \[\sqrt{a*b*c} = \sqrt{a}*\sqrt{b}*\sqrt{c}\]

  14. miah
    • 2 years ago
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    Im so confused/:

  15. SandeepReddy
    • 2 years ago
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    \[\sqrt{2*8^2*(x^2)^2*x*y^2}\] = \[\sqrt{2}*\sqrt{8^2}*\sqrt{(x^2)^2}*\sqrt{x}*\sqrt{y^2}\]

  16. SandeepReddy
    • 2 years ago
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    that gives u \[\sqrt{2}*8*x^2*\sqrt{x}*y\]

  17. SandeepReddy
    • 2 years ago
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    \[8\sqrt{2}x^{3/2}y\]

  18. SandeepReddy
    • 2 years ago
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    @miah did u get it?

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

  20. SandeepReddy
    • 2 years ago
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    Yes i too feel same @aajugdar

  21. miah
    • 2 years ago
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    Yes it would have been easier. But i some what get it.

  22. aajugdar
    • 2 years ago
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    smwhat is better than nothing

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