How do you rationalize the denominator and simplify #(5sqrt6)/sqrt10#?

3 Answers
Mar 9, 2018

Answer:

#sqrt15#

Explanation:

multiply by #sqrt10#:

#(5sqrt6sqrt10)/((sqrt10)^2)#

#sqrt6 * sqrt10 = sqrt (6*10) = sqrt60#

#(sqrt10)^2 = 10#

#(5sqrt6)/(sqrt10) = (5sqrt60)/10#

#sqrt60 = sqrt4 * sqrt15 = 2sqrt15#

#5sqrt60 = 5 * 2 * sqrt15 = 10sqrt15#

#(5sqrt60)/10 = (10sqrt15)/10#

#= (sqrt15)/1#

#= sqrt15#

Mar 9, 2018

Answer:

#sqrt15#

Explanation:

In order to rationalize the denominator, you can multiply by #sqrt10/sqrt10#. This is the same as multiplying the fraction by one. If you multiply #(5sqrt6)/sqrt10 *sqrt10/sqrt10#, you get #(5sqrt60)/10#. If you multiply the square roots in the denominator, you get #sqrt100#, which is equivalent to 10.

With #(5sqrt60)/10#, you can simplify to just #sqrt60/2#.

Next, you can simplify #sqrt60# by doing a factor tree. When you do a factor tree, you will find you can pull out a factor of 2 from the square root leaving you with #(2sqrt15)/2#.

Lastly, just cancel out the 2 in the numerator and denominator, and you get the answer of #sqrt15#

Mar 9, 2018

Answer:

#sqrt15#

Explanation:

#"using the "color(blue)"laws of radicals"#

#•color(white)(x)sqrtaxxsqrtbhArrsqrtab#

#•color(white)(x)sqrtaxxsqrta=a#

#"To rationalise the denominator that is eliminate the "#
#"radical from the denominator"#

#"multiply numerator/denominator by "sqrt10#

#rArr(5sqrt6)/sqrt10#

#=(5xxsqrt6xxsqrt10)/(sqrt10xxsqrt10)#

#=(5xxsqrt60)/10#

#=(5xxsqrt(4xx15))/10#

#=(5xxsqrt4xxsqrt15)/10#

#=(5xx2xxsqrt15)/10=(cancel(10)sqrt15)/cancel(10)=sqrt15#