How do you simplify #\frac { \sqrt { 2- \sqrt { 3} } } { \sqrt { 2+ \sqrt { 3} } }#?

2 Answers
IDKwhatName
Aug 9, 2017

#-(13-4sqrt(3))/13#

Explanation:

Square top and bottom, to remove both square roots:
#(sqrt(2-sqrt(3))/sqrt(2+sqrt(3)))^2=(sqrt(2-sqrt(3)))^2/(sqrt(2+sqrt(3)))^2=(2-sqrt(3))/(2+sqrt(3))#

Now, to rationalise the denominator by multiplying the top and bottom by the negative of the square root:
#((2-sqrt(3))(2-sqrt(3)))/((2+sqrt(3))(2-sqrt(3)))=(4-2sqrt(3)-2sqrt(3)+9)/(4+2sqrt(3)-2sqrt(3)-9)=(13-4sqrt(3))/(-13)=-(13-4sqrt(3))/13#

iceman
Aug 9, 2017

#= (2 - sqrt3)#

Explanation:

#sqrt(2 - sqrt3)/sqrt(2+sqrt3)# => Rationalize the denominator:

#= sqrt(2 - sqrt3)/sqrt(2+sqrt3) * sqrt(2 - sqrt3)/sqrt(2-sqrt3)#

#= (2 - sqrt3)/sqrt(4-sqrt9)#

#= (2 - sqrt3)/sqrt(4 - 3)#

#= (2 - sqrt3)/sqrt1#

#= (2 - sqrt3)#

Related questions
Impact of this question
2369 views around the world
You can reuse this answer
Creative Commons License