$\displaystyle

\sqrt{8x^5y^6} = \sqrt{4x^4y^62x}

$

why $\displaystyle 2x$?

why not

$\displaystyle \sqrt{8x^5y^6} = \sqrt{8x^4y^6x}$

?

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- Feb 12th 2011, 12:32 AMalyosha2Simplifying radicals
$\displaystyle

\sqrt{8x^5y^6} = \sqrt{4x^4y^62x}

$

why $\displaystyle 2x$?

why not

$\displaystyle \sqrt{8x^5y^6} = \sqrt{8x^4y^6x}$

? - Feb 12th 2011, 12:34 AMMoo
Because 4 is a square while 8 isn't :)

- Feb 12th 2011, 12:57 AMalyosha2
I mean why factor the 8 out at this point? The odd exponent has been factored out of the x to make it even. I understand that at some point the 8 will have to be factored, but is it just pure subjctive preference that it has been pulled out here rather than doing it later? or does it have some thing to do with making the x's exponent even?

- Feb 12th 2011, 01:11 AMProve It
Because you're looking for square factors, so that you can take square roots...