How do you simplify # sqrt(27k^7q^8)#?

1 Answer
MathFact-orials.blogspot.com
Jun 8, 2016

There are different ways to look at how to simplify root statements - I present one of mine below and get to #3k^3q^4root(2)(3k)#

Explanation:

When we're dealing with simplifying a term within a root, I like to use the "buddy" rule - which is, in order for a term to "escape" the root, it needs the number of buddies listed in the root (a square root is 2 buddies, a cube root is 3 buddies, etc). When there is no number, like in this case, the default is 2.

Let's start with the original (but I'm going to add the 2 to indicate clearly this is a square root):

#root(2)(27k^7q^8)#

Now let's break down each part of the terms inside the root piece by piece:

27

#27=(3*3)*3#, which means there is one of 3 "buddies" and one last 3. So one of the 3 "buddies" gets out and the last 3 stays inside the root.

#k^7#

#k^7=(k*k)*(k*k)*(k*k)*k#, which means there are 3 sets of k "buddies" so 3 k's get out from under the root and the last k is left inside the root.

#q^8#

#q^8=(q*q)*(q*q)*(q*q)*(q*q)#, which means there are 4 sets of q "buddies" so 4 q's escape the root and none stay inside the root.

Putting it all together, we get:

#3k^3q^4root(2)(3k)#

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