Question

How do you simplify `33/sqrt99`?

Answer 1

`sqrt11`

Explanation:

`sqrt99=sqrt(9xx11)=sqrt9xxsqrt11=3sqrt11`

`rArr(33)/(sqrt99)=cancel(33)^(11)/(cancel(3)^1sqrt11)=11/sqrt11`

`color(blue)" Rationalising the denominator"`

`rArr11/sqrt11xxsqrt11/sqrt11`

`=(cancel(11)^1sqrt11)/cancel(11)^1`

`=sqrt11`

Answer 2

We can also do this in one fell swoop using prime factorization:

`33/sqrt(99)=(3xx11)/(3^2xx11)^(1/2)=(3xx11)/(3xx11^(1/2))=11^(1-1/2)=11^(1/2)=sqrt11`