How do you simplify #(4x)^(1/2)*(9x)^(1/2)#?

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Answer 1 · 2015-06-14T19:55:56

Use the rule #x^(m/n)=root(n)(x^m)# #; n!=0#
http://www.regentsprep.org/regents/math/algtrig/ato1/fractionalexp.htm
Solve as usual to get #6x#.

#x^(m/n)=root(n)(x^m)# #; n=2="square root"# #;m=1#

#(4x)^(1/2)*(9x)^(1/2)# =

#sqrt(4x)*sqrt(9x)# =

#2sqrt x*3sqrtx# =

#2*3sqrt xsqrtx# =

#6sqrt(x^2)# =

#6x#