How do you simplify #sqrt(216t)+sqrt(96t)#?

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Answer 1 · 2016-12-26T19:03:30

#=10sqrt(6t)#

Find the prime factors under each root - then you know what you are working with...

#sqrt(2^3 xx 3^3 xx t) + sqrt(2^5 xx 3xxt)" "larr# make even indices

#=sqrt(2^2 xx 2 xx3^2xx3 xxt) + sqrt(2^4 xx 2xx3xxt)#

#=2xx3sqrt(2xx3xxt)+2^2sqrt(2xx3xxt)" "larr# find possible roots

#=2xx3sqrt(6t)+2xx2sqrt(6t)" "larr# factor out the HCF

#=2sqrt(6t)(3+2)#

#=2xx5sqrt(6t)#

#=10sqrt(6t)#