Question

How do you simplify `\frac { 9} { \sqrt { 21} }`?

Answer 1

`27/7`

Explanation:

Firstly, we have to get rid of the radical. To do that, we can `^2`(square) the entire fraction and then solve from there.

`(9/sqrt21)^2=(9*9)/(sqrt21)^2=81/21` ~(3 can be taken out)~ `=27/7`

Answer 2

`(3sqrt21)/7`

Explanation:

To eliminate the `sqrt21` from the denominator of the fraction we use a method called `color(blue)"rationalising"`

This ensures that we have a `color(blue)"rational"` denominator as opposed to a `color(blue)"surd".`

Consider `sqrt100xxsqrt100`

`sqrt100=10`

`rArrsqrt100xxsqrt100`

`=10xx10`

`=100larrcolor(red)" the value inside the root"`

`"in general " sqrtaxxsqrta=a`

`"Since " 9/sqrt21" is a fraction"` we multiply the numerator/denominator by `sqrt21`

`rArr9/sqrt21xxsqrt21/sqrt21`

`=(9xxsqrt21)/21larrcolor(red)" using above result"`

`=(cancel(9)^3xxsqrt21)/cancel(21)^7larrcolor(red)" cancelling by 3"`

`=(3sqrt21)/7`

Answer 3

`color(red)(=(3sqrt21)/7`

Explanation:

`9/sqrt21`

`:.=3^2/(sqrt3 xx sqrt7)`

`:.=3^2/(3^(1/2)sqrt7)`

`:.=3^(2-1/2)/sqrt7`

`:.=(3^(1 1/2))/(sqrt7)`

`:.=3^(3/2)/sqrt7`

`:.=root2(3^3)/sqrt7`

`:.=sqrt(3^3)/sqrt7`

Rationalize denominator by multiplying by `sqrt7/sqrt7`

`:.=(sqrt(3^3))/sqrt7 xx sqrt7/sqrt7`

`:.=sqrt3*sqrt3=3,sqrt7*sqrt7=7`

`:.=sqrt(3*3*3*7)/7`

`:.=(3sqrt(3*7))/7`

`:.color(red)(=(3sqrt21)/7`