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
#root(3)(x^2xxy^3)#

Our first task is to expand all our components

#root(3)(x xx x xxy xx y xx y)#

Now we need to look at our index, the #3# in the top left corner

Whatever that number is, that's the value of the exponent we're trying to create. If it's a #sqrt(color(white)(x))#, we want an #x^2#. If it's a #root(3)(color(white)(x))#, we want a #x^3#.

So, we are looking for #x^3# inside the root, or #x xx x xx x#.
Back to our example:

Let's condense everything we can to give us a cube root

#root(3)(x xx x xx y^3)#

When we have a number that matches the index, we can take it out of the root:

#yroot(3)(x xx x)#

We can simplify that to #yroot(3)sqrt(x^2)#

#color(white)(0)#

Now, let's solve your problem:

#sqrt(108)#

expand

#sqrt(2 xx 3 xx 3 xx 3 xx 2)#

what's the index?
#2#
What can we change to something squared

#sqrt(2^2 xx 3^2 xx 3)#

Bring out the squared

That gives us #2xx3sqrt(3)# or #6sqrt(3)#

#color(white)(0)#

Let's look at the other option

#2sqrt(27)#(#sqrt(4*27)#)

#2sqrt(4*27)# equals #sqrt(108)#

#2sqrt(27)*xxsqrt(108)#

#2sqrt(3xx3xx3)xxsqrt(2xx2xx3xx3xx3)#

#2sqrt(3^2xx3)xxsqrt(2^2xx3^2xx3)#

#2xx3sqrt(3)xx6sqrt(3)#

#36sqrt(3xx3)#

#36xx3#

#108#

#108# does not equal #sqrt(108)#, so the second option cannot be correct. Hopefully this helps!

May 17, 2017

Answer:

#sqrt(108) = 6sqrt(3) = 3sqrt(12) = 2sqrt(27)#

the "simplest" being #6sqrt(3)#.

Explanation:

When #a >=0# and/or #b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

Also, if #a >= 0# then:

#sqrt(a^2) = a#

The prime factorisation of #108# is:

#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 #sqrt(108)#.

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, #6sqrt(3)# is considered simplified, whereas #2sqrt(27)# or #3sqrt(12)# are not fully simplified.

So if your teacher marked #2sqrt(27)# (from #sqrt(108) = sqrt(4*27)#) as being wrong, it was not because it was not equal (it is), but because it was not fully simplified.