How do you simplify #(12sqrt9)/sqrt18#?

Question

931 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T17:17:58

# = 6sqrt2 #

#(12sqrt9)/ sqrt18#

#sqrt 9 = sqrt ( 3 *3 ) = color(green)(3 #

#sqrt18 = sqrt ( 2 * 3 * 3 ) = sqrt ( 2 * 3 ^2) = color(green)( 3 sqrt2 #

Our expression now becomes
#(12 * color(green)((3)))/ color(green)( 3 sqrt2 #

# = (12 * color(green)(cancel(3)))/ color(green)( cancel3 * sqrt2 #

# = 12 / sqrt2 #

Rationalising the denominator:
# = (12 * sqrt2) / (sqrt2 * sqrt2) #

# = (12 * sqrt2) / 2 #

# = (cancel12 * sqrt2) / cancel 2 #

# = 6sqrt2 #