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

2 Answers
Apr 23, 2017

Answer:

#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#

Apr 23, 2017

Answer:

#(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#