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

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Answer 1 · 2015-04-03T01:17:37

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

$(5sqrt(6))/sqrt(10)$

$= (5sqrt(6))/(sqrt(10))*(sqrt(10))/(sqrt(10))$

$=(5sqrt(60))/(10)$

which can be simplified as
$(5*(2)sqrt(15))/(10)$

$= sqrt(15)$

Answer 2 · 2015-04-03T01:26:46

Two methods:

Method 1

Multiply by $1$ in the form $sqrt 10 / sqrt 10$ to get:

$(5sqrt6)/sqrt10 * sqrt10/sqrt10 = (5sqrt 60) /10$

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

$(5sqrt6)/sqrt10 = (5sqrt 60) /10 = (5*2 sqrt 15)/10 = sqrt 15$

Method 2

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

$(5sqrt6)/sqrt10=(5sqrt3sqrt2)/(sqrt5sqrt2)=(5sqrt3)/sqrt5$

Method 2a: now multiply by $1$ in the form $sqrt5/sqrt5$ to get:

$(5sqrt6)/sqrt10= (5sqrt3)/sqrt5*sqrt5/sqrt5 = (5sqrt15)/5 = sqrt15$

Method 2b Reduce $a/sqrta = sqrta$ , So $5/sqrt5 = sqrt5$
*
$(5sqrt6)/sqrt10= (5sqrt3)/sqrt5 = ((sqrt5)^2sqrt3)/sqrt5=sqrt5sqrt3= sqrt15$

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