How do you simplify #sqrt(8u^8 v^3) sqrt(2u^5 v^6)#?

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-02T23:29:08

#sqrt(8u^8v^3)sqrt(2u^5v^6) = 4u^6v^4sqrt(uv)#

with restriction #u, v >= 0#

If #sqrt(8u^8v^3)# is defined then #8u^8v^3 >= 0#

If #sqrt(2u^5v^6)# is defined then #2u^5v^6 >= 0#

Use #sqrt(a)sqrt(b) = sqrt(ab)# when #a, b >= 0# to get:

#sqrt(8u^8v^3)sqrt(2u^5v^6)#

#=sqrt(8u^8v^3*2u^5v^6)#

#=sqrt(16u^13v^9)#

#=sqrt((4u^6v^4)^2uv)#

#= 4u^6v^4sqrt(uv)#