How do you simplify #sqrt((1/18))#?

2 Answers
Oct 30, 2015

Answer:

#sqrt(2)/6#

Explanation:

Since there can't be a square root (a radical) in the denominator, rationalize it by multiplying both the numerator and denominator with #sqrt(2)#.

#sqrt(1/18) * sqrt(2)/sqrt(2)#

#=sqrt(2/36)#

#=1/6sqrt(2)# or #sqrt(2)/6#

You multiply by #sqrt(2)# in order to make the denominator #sqrt(36)# because #sqrt(36)=6# and so you can take it out of the square root.

Oct 30, 2015

Answer:

#sqrt(2)/6#

Explanation:

It is a matter of splitting the numbers up into factors that have a root if you can. Then taking these outside of the root by applying that root.

What factors are there of 18 that we can apply a root to?
The obvious ones are 2 and 9 as #2 times 9 =18# we can take the root of 9 but not of 2. So we end up with:

#sqrt((1/18)) = sqrt(1/(2 times 9)#

This can be split so that we have:

#sqrt(1/2 times 1/9) = 1/3 sqrt(1/2)#

But convention is that you do not have a root as a denominator if you can help it.

Write as: #1/3 (sqrt(1))/( sqrt(2))# This does work. Check it on a calculator.

But #sqrt(1) =1# giving:

#1/3 times 1/(sqrt(2))#

To 'get rid' of the root in the denominator multiply by the value 1 (does not change the overall values) but write the 1 in the form of #(sqrt(2))/(sqrt(2))# giving:

#1/3 times 1/sqrt(2) times sqrt(2)/sqrt(2)#

But #sqrt(2) times sqrt(2) = 2#

So now we have:
#1/3 sqrt(2)/2 = sqrt(2)/6#

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