How do you simplify #\sqrt { 225x ^ { 18} y ^ { 22} }#?

Question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-11T04:35:46

#15x^9y^11#

Recall : #sqrtx=x^(1/2), and sqrt(x^2)=(x^2)^(1/2)=x^(2*1/2)=x#,

#sqrt(225x^18y^22)=sqrt(15*15*x^18*y^22)=(15^2*x^18*y^22)^(1/2)#
#=15^(2*1/2)*x^(18*1/2)*y^(22*1/2)#
#=15*x^9*y^11#

Answer 2 · 2016-12-11T09:07:09

#sqrt(225x^18y^22) = abs(15x^9y^11)#

Note that:

#225x^18y^22 = 15^2*(x^9)^2*(y^11)^2 = (15x^9y^11)^2#

So #15x^9y^11# is a square root of #225x^18y^22#.

We also find:

#(-15x^9y^11)^2 = 225x^18y^22#

So #-15x^9y^11# is also a square root of #225x^18y^22#.

Note that if #a >= 0# then #sqrt(a)# denotes the non-negative square root.

So if, for example, #x < 0# and #y > 0# then #15x^9y^11 < 0# would be the wrong square root.

We can express that we want the non-negative square root by taking the modulus:

#sqrt(225x^18y^22) = abs(15x^9y^11)#