How do you simplify #sqrt(24 x^10 y^11)#?

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Answer 1 · 2016-08-19T15:58:34

#2x^5y^5sqrt(6y)#

Note that #24" " =" " 4xx6" "=" " 2^2xx6# giving:

#sqrt(2^2xx6x^10y^11)#

take the #2^2# outside the square root.

#2sqrt(6x^10y^11)#

Note that #sqrt(x^10)" "=sqrt(x^2xx x^2 xx x^2xx x^2xx x^2)=x^5#

Take all of the #x^2# outside the square root.

#2x^5sqrt(6y^11)#

Note that #sqrt(y^11)" " =" " sqrt((y^2)^5xxy)" " =" " y^5sqrt(y)#

#2x^5y^5sqrt(6y)#