How do you simplify #sqrt(108)# ?
2 Answers
Well, first, let's explain how you simplify a square root:
Let's say the problem is this
Our first task is to expand all our components
Now we need to look at our index, the
Whatever that number is, that's the value of the exponent we're trying to create. If it's a
So, we are looking for
Back to our example:
Let's condense everything we can to give us a cube root
When we have a number that matches the index, we can take it out of the root:
We can simplify that to
Now, let's solve your problem:
expand
what's the index?
What can we change to something squared
Bring out the squared
That gives us
Let's look at the other option
the "simplest" being
Explanation:
When
#sqrt(ab) = sqrt(a)sqrt(b)#
Also, if
#sqrt(a^2) = a#
The prime factorisation of
#108 = 2*2*3*3*3#
So we find:
#sqrt(108) = sqrt(2^2*27) = sqrt(2^2)sqrt(27) = 2sqrt(27)#
#sqrt(108) = sqrt(3^2*12) = sqrt(3^2)sqrt(12) = 3sqrt(12)#
#sqrt(108) = sqrt(6^2*3) = sqrt(6^2)sqrt(3) = 6sqrt(3)#
So there are several "simplifications" of
When asked to simplify such a square root, what is normally expected is the expression that no longer has any square factors in the radicand. So in our example,
So if your teacher marked