How do you rationalize the denominator (5 sqrt 6)/(sqrt 10)5√6√10?

2 Answers
Alan P.
Apr 3, 2015

If you multiply both the numerator and the denominator by sqrt(10)√10
then the denominator will become a non-radical.

(5sqrt(6))/sqrt(10)5√6√10

= (5sqrt(6))/(sqrt(10))*(sqrt(10))/(sqrt(10))=5√6√10⋅√10√10

=(5sqrt(60))/(10)=5√6010

which can be simplified as
(5*(2)sqrt(15))/(10)5⋅(2)√1510

= sqrt(15)=√15

Jim H
Apr 3, 2015

Two methods:

Method 1

Multiply by 11 in the form sqrt 10 / sqrt 10√10√10 to get:

(5sqrt6)/sqrt10 * sqrt10/sqrt10 = (5sqrt 60) /105√6√10⋅√10√10=5√6010

Now simplify sqrt 60 = sqrt (4*15) = 2 sqrt 15√60=√4⋅15=2√15 So we have

(5sqrt6)/sqrt10 = (5sqrt 60) /10 = (5*2 sqrt 15)/10 = sqrt 155√6√10=5√6010=5⋅2√1510=√15

Method 2

Notice that 6 and 10 are both divisible by 2, so we can start by simplifying:

(5sqrt6)/sqrt10=(5sqrt3sqrt2)/(sqrt5sqrt2)=(5sqrt3)/sqrt55√6√10=5√3√2√5√2=5√3√5

Method 2a: now multiply by 11 in the form sqrt5/sqrt5√5√5 to get:

(5sqrt6)/sqrt10= (5sqrt3)/sqrt5*sqrt5/sqrt5 = (5sqrt15)/5 = sqrt155√6√10=5√3√5⋅√5√5=5√155=√15

Method 2b Reduce a/sqrta = sqrtaa√a=√a , So 5/sqrt5 = sqrt55√5=√5
*
(5sqrt6)/sqrt10= (5sqrt3)/sqrt5 = ((sqrt5)^2sqrt3)/sqrt5=sqrt5sqrt3= sqrt155√6√10=5√3√5=(√5)2√3√5=√5√3=√15

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