How do you solve #sqrt(243x^8y^5)#?

Question

1269 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-10T15:45:41

#= 9 x^4 y^2sqrt ( 3y #

#sqrt (243 x^8 y^5#

Simplifying #sqrt243# by prime factorisation:
#243 = 3*3*3*3*3 = 3^5#

#sqrt (243 x^8 y^5) = sqrt (color(green)(3^5) x^8 y^5 #

#= sqrt (color(green)(3^4* 3) * x^8 * y^4 *y #

( note: Square root , can also be called as second root , so in terms of fraction, second root is a half power #color(blue)(1/2# )

so, #sqrt (3^4) = 3^2#, #sqrt(x^8)= x^4# and #sqrt(y^4) = y^2#

#sqrt (color(green)(3^4* 3) * x^8 * y^4 * y ) = 3^2 * x^4 * y^2sqrt (color(green)( 3) y#

#= 9 x^4 y^2sqrt ( 3 y #