How do you simplify #(sqrt9sqrt8)/(sqrt6sqrt6)?#

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Answer 1 · 2017-03-30T11:38:37

#sqrt2#

#(sqrt9sqrt8)/(sqrt6sqrt6)=(3*sqrt(2*4))/6#

#=(3*2sqrt(2))/6=(6sqrt2)/6#

#=sqrt2#

Answer 2 · 2017-03-30T11:45:38

There are different methods which can be used:

First option: Combine the radicals:

#(sqrt9sqrt8)/(sqrt6sqrt6) = sqrt72/sqrt36 = sqrt(72/36) = sqrt2#

Second option: Simplify where possible, find factors of radicands.

#(color(red)(sqrt9)color(blue)(sqrt8))/(color(green)(sqrt6sqrt6))#

#=(color(red)(3)color(blue)(sqrt(4xx2)))/(color(green)(sqrt6)^2)#

#=(color(red)(3)color(blue)(xx2sqrt(2)))/(color(green)(6)#

#=sqrt2#

Third option: Write as the product of prime factors:

#(sqrt9sqrt8)/(sqrt6sqrt6)#

#=(sqrt3 xx sqrt3 xxsqrt2xx sqrt2xx sqrt2)/(sqrt2xxsqrt3xxsqrt2xx sqrt3)" "# now cancel

#=(cancelsqrt3 xx cancelsqrt3 xxcancelsqrt2xx cancelsqrt2xx sqrt2)/(cancelsqrt2xxcancelsqrt3xxcancelsqrt2xx cancelsqrt3)#

#=sqrt2#