Question

How do you simplify `sqrt(3/16)*sqrt(9/5)`?

Answer 1

`sqrt(3/16)*sqrt(9/5) = (3sqrt(15))/20`

Explanation:

Note that if `a, b >= 0` then:

`sqrt(ab) = sqrt(a)sqrt(b)`

If `a > 0` and `b >= 0` then:

`sqrt(a/b) = sqrt(a)/sqrt(b)`

If `a >= 0` then:

`sqrt(a^2) = a`

When simplifying square roots of rational expressions, I like to make the denominator square before splitting the square root. That way we don't have to rationalise the denominator later...

`sqrt(3/16)*sqrt(9/5) = sqrt(3/16)*sqrt((9*5)/(5*5))`

`color(white)(sqrt(3/16)*sqrt(9/5)) = sqrt(3/4^2)*sqrt(45/5^2)`

`color(white)(sqrt(3/16)*sqrt(9/5)) = sqrt((3*45)/(4^2*5^2))`

`color(white)(sqrt(3/16)*sqrt(9/5)) = sqrt(3*45)/sqrt(4^2*5^2)`

`color(white)(sqrt(3/16)*sqrt(9/5)) = sqrt(3*3^2*5)/sqrt((4*5)^2)`

`color(white)(sqrt(3/16)*sqrt(9/5)) = (sqrt(3^2)*sqrt(15))/sqrt(20^2)`

`color(white)(sqrt(3/16)*sqrt(9/5)) = (3sqrt(15))/20`

Answer 2

`(3sqrt15)/20`

Explanation:

`sqrt(3/16)*sqrt(9/5)`

`:.=sqrt3/sqrt16*sqrt9/sqrt5`

`:.=sqrt3/sqrt(4*4)*sqrt(3*3)/sqrt5`

`:.sqrt4*sqrt4=4,sqrt3*sqrt3=3`

`:.=sqrt3/4*3/sqrt5`

`:.=(3sqrt3)/(4sqrt5)*sqrt5/sqrt5`

`:.=sqrt5*sqrt5=5`

`:.=(3sqrt15)/(4*5)`

`:.=(3sqrt15)/20`