How do you write #sqrt(24a^10b^6)# in simplest form?

3 Answers
Gió
Mar 24, 2015

You can write it as:
#sqrt(6*4*a^(10)b^6)=2a^5b^3sqrt(6)#

Where you used the fact that #sqrt(x)=x^(1/2)#

and so:
#sqrt(6*4*a^(10)b^6)=(2^2*6*a^(10)b^6)^(1/2)=#
#=2^(2*1/2)6^(1/2)a^(10*1/2)b^(6*1/2)=2a^5b^3sqrt(6)#

Jim H
Mar 24, 2015

If we may assume that #a# and #b# are non-negative, then we write #sqrt(24a^10b^6)=2a^5b^3sqrt6#

#sqrt(24a^10b^6)=sqrt(4*6*(a^5)^2(b^3)^2)=2sqrt6a^5a^3=2a^5b^3sqrt6#

The key is the repeated use of: #sqrt(n^2)=n#
(for non-negative #n#).

So #sqrt4=sqrt(2^2)=2#, and #sqrt((a^5)^2)=a^5# and #sqrt((b^3)^2)=b^3#.

(If we may NOT assume #a# and #b# are non-negative, the we need absolute values, because then we would need #sqrt(a^2)=absa#
#sqrt(24a^10b^6)=2abs(a^5b^3)sqrt6#.)

Alan P.
Mar 24, 2015

Extract the largest square from each of the terms within the square root:
#24 = 2^2 * 6#
#a^(10) = (a^5)^2 *(1)#
#b^6 = (b^3)^2 *(1)#

So
#sqrt(24 a^(10)b^6)#

#=sqrt( (color(red)(2)^2)(6)color(red)((a^5))^2color(red)((b^3))^2)#

#= color(red)(2a^5b^3) sqrt(6)#

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