How do you simplify #\sqrt { 225x ^ { 3} y ^ { 9} } #?

1 Answer
May 16, 2018

#15xy^4sqrt(xy)#

Explanation:

We must find as many squares as possible inside the square root: for the numeric part, we have #225 = 15^2#, which is already a square. For the variable part, we have #x^3 = x^2\cdot x#, and #y^9=y^8\cdot y#. Note that every variable raised to an even power is a square, since #x^(2k) = (x^k)^2#.

Now, use the fact that the square root of a multiplication is the multiplications of the square roots: #sqrt(ab)=sqrt(a)sqrt(b)# to get

#sqrt(225x^3y^9) = sqrt(15^2\cdot x^2\cdot x \cdoty^8\cdot y) = sqrt(255)sqrt(x^2)sqrt(x)sqrt(y^8)sqrt(y)#

Simplify the squares with the square roots:

#sqrt(255)sqrt(x^2)sqrt(x)sqrt(y^8)sqrt(y) = 15x\sqrt(x)y^4sqrt(y)#

Rearrange for better visualization:

#15x\sqrt(x)y^4sqrt(y) = 15xy^4sqrt(xy)#