Rationalise Denominator: 1/(1 + √2 + √3) by using: (1 - √2 + √3) as conjugate?

2 Answers
Kalit Gautam
Feb 24, 2018

Explained below :-

Explanation:

#k = 1/(1 + {√2 + √3})#

#rArr k = 1/({√3+1}+√2)#

#rArrk = ({√3+1}-√2)/[({√3+1}+√2).({√3+1}-√2)]#

#rArrk = ({√3+1}-√2)/[(√3+1)^2-(√2)^2]#

#rArrk = ({√3+1}-√2)/[3+1+2√3-2]#

#:.k = ({√3+1}-√2)/[2(1+√3)]#

George C.
Mar 6, 2018

#1/(1+sqrt(2)+sqrt(3))=(2+sqrt(2)-sqrt(6))/4#

Explanation:

The difference of squares identity can be written:

#A^2-B^2=(A-B)(A+B)#

This is the key to eliminating square roots from the denominator.

Note that #(1-sqrt(2)+sqrt(3))# is only a partial conjugate for #(1+sqrt(2)+sqrt(3))#. Multiplying these two expressions will eliminate terms in #sqrt(2)# but leave terms in #sqrt(3)#.

If we want to rationalise the denominator, we will also need to multiply by some expression of the form #a+bsqrt(3)# (actually I will use #(sqrt(3)-1)#)

#1/(1+sqrt(2)+sqrt(3))#

#=((1+sqrt(3))-sqrt(2))/(((1+sqrt(3))-sqrt(2))((1+sqrt(3))+sqrt(2)))#

#=(1+sqrt(3)-sqrt(2))/((1+sqrt(3))^2-(sqrt(2))^2)#

#=(1+sqrt(3)-sqrt(2))/(1+2sqrt(3)+3-2)#

#=(1+sqrt(3)-sqrt(2))/(2(sqrt(3)+1))#

#=((sqrt(3)-1)(1+sqrt(3)-sqrt(2)))/(2(sqrt(3)-1)(sqrt(3)+1))#

#=((sqrt(3)+3-sqrt(6))-(1+sqrt(3)-sqrt(2)))/(2(3-1))#

#=(2+sqrt(2)-sqrt(6))/4#

Note that having got to this result, the numerator is a proper conjugate for #1+sqrt(2)+sqrt(3)# in that:

#(1+sqrt(2)+sqrt(3))(2+sqrt(2)-sqrt(6)) = 4#

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