How do you rationalize the denominator and simplify #(3sqrt9)/sqrt27#?
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You can simplify the expression quite a lot before rationalizing:
- #sqrt(9)=sqrt(3^2)=3#
- #sqrt(27)=sqrt(9*3)=sqrt(9)*sqrt(3)=3sqrt(3)#
So, the expression becomes
#\frac{cancel(3)*3}{cancel(3)sqrt(3)}#
To rationalize this expression, multiply and divide by #sqrt(3)#:
#\frac{3}{sqrt(3)} = \frac{3}{sqrt(3)} * \frac{sqrt(3)}{sqrt(3)} = \frac{3sqrt(3)}{sqrt(3)sqrt(3)} = \frac{cancel(3)sqrt(3)}{cancel(3)} = sqrt(3)#
#(3sqrt9)/sqrt27=(3*3)/(3sqrt3)=3/sqrt3=(3sqrt3)/3=sqrt3#
#(3 sqrt 9)/sqrt 27#
#:.=(3 sqrt 9)/sqrt 27 xx (sqrt 27)/ (sqrt 27)#
#sqrt27*sqrt27=27#
#:.=(3 sqrt(9*27))/27#
#:.=(3 sqrt((3*3)(3*3*3)))/27#
#sqrt3*sqrt 3=3#
#:.=(3*3*3 sqrt 3)/27#
#:.=(cancel27^1 sqrt 3)/cancel27^1#
#:.=sqrt 3#
First, we can rewrite #color(blue)(sqrt27)# as #sqrt3*sqrt9#, which is the same as #3sqrt3#. Doing this leaves us with
#(3sqrt9)/(color(blue)(3sqrt3))#
A #3# in the numerator and denominator cancels, leaving us with
#sqrt9/sqrt3#, which is also equal to
#3/sqrt3#
The key realization is that when we rationalize the denominator, we multiply the top and bottom by that denominator.
In our example, we have
#(3*sqrt3)/(sqrt3*sqrt3)#
Since we are multiplying by #1# essentially, we are not changing the expression's value. The denominator simplifies to #3#, and we're left with
#sqrt3#
Even though we could have rationalized the denominator in the intermediate steps, it never hurts to simplify first.
Hope this helps!
