How do you simplify #(sqrt5)/(sqrt5-sqrt3)#?

3 Answers
Apr 19, 2018

Answer:

#(5 + sqrt(15))/2#

Explanation:

#=> sqrt(5)/(sqrt(5) - sqrt(3))#

Multiply and divide by #(sqrt(5) + sqrt(3))#

#=> sqrt(5)/(sqrt(5) - sqrt(3)) × (sqrt(5) + sqrt(3))/(sqrt(5) + sqrt(3))#

#=> (sqrt(5)(sqrt(5) + sqrt(3)))/((sqrt(5) - sqrt(3))(sqrt(5) + sqrt(3))#

#=> (sqrt(5)(sqrt(5) + sqrt(3)))/((sqrt(5))^2 - (sqrt(3))^2) color(white)(..)[∵ (a - b)(a + b) = a^2 - b^2]#

#=> (sqrt(5)sqrt(5) + sqrt(5)sqrt(3))/(5 - 3)#

#=> (5 + sqrt(15))/2#

Apr 19, 2018

Answer:

#(5+sqrt(15)) / 2 #

Explanation:

Multiply #(√5) / (√5−√3)# by #(√5+√3) / (√5+√3)# to rationalize the denominator

#(√5)/(√5−√3)# * #(√5+√3) / (√5+√3)# = #(sqrt5*(sqrt5 + sqrt3)) / 2#

Apply the distributive property

#(sqrt5*(sqrt5 + sqrt3)) / 2# = #((sqrt5*sqrt5)+(sqrt5*sqrt3))/2# = #(5+sqrt(15)) / 2 #

Apr 19, 2018

Answer:

# = 5/(5 - (sqrt(15))#
OR
#= 5/2 + sqrt(15)/2#
Take your pick.

Explanation:

These days, it may be simplest to just use a calculator to complete the expression. But, for purposes of demonstration, we multiply by a radical factor just as we would with another number.
#sqrt(5)/(sqrt(5) - sqrt(3)) xx sqrt(5)/(sqrt(5)# # = 5/(5 - (sqrt(3) xx sqrt(5))#

#5/(5 - (sqrt(3) xx sqrt(5))## = 5/(5 - (sqrt(15))#

OR
Multiply the denominator and numerator by the same expression as the denominator but with the opposite sign in the middle. This expression is called the conjugate of the denominator.

#sqrt(5)/(sqrt(5) - sqrt(3)) xx (sqrt(5) + sqrt(3))/(sqrt(5) + sqrt(3))#

#= (5 + sqrt(15))/(5 - 3)# = #(5 + sqrt(15))/2 = 5/2 + sqrt(15)/2#

https://www.mathportal.org/algebra/roots-and-radicals/multiplying-and-dividing-radicals.php