How do you prove that #sqrt(4+2sqrt(3)) = sqrt(3)+1# ?

One way to show that the left hand side is equal to the right hand side is to show that their quotient is equal to #1#. Beginning with the quotient, we have

#sqrt(4+2sqrt(3))/(sqrt(3)+1)#

To help us evaluate this, let's first rationalize the denominator

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

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

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

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

As the quotient must be equal to #1# if the given expressions are equal, we now need to show that the numerator is equal to #2#.

#sqrt(4+2sqrt(3))xx(sqrt(3)-1) = sqrt(4+2sqrt(3))xxsqrt((sqrt(3)-1)^2)#

(Note that the above step is justified because #sqrt(3)-1 > 0#. If #x>=0#, then #x = sqrt(x^2)#. If #x<0#, then #x=-sqrt(x^2)#)

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

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

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

#=sqrt(4^2-(2sqrt(3))^2)#

#=sqrt(16-12)#

(As when rationalizing the denominator, we make use of the identity #(a+b)(a-b) = a^2-b^2#)

#=sqrt(4)#

#=2#

Now that we have shown that the numerator has the desired property, we can solve the rest of the problem quite simply.

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

#=> sqrt(4+2sqrt(3))/(sqrt(3)+1)xx(sqrt(3)+1) = 1xx(sqrt(3)+1)#

#:. sqrt(4+2sqrt(3)) = sqrt(3)+1#

This expression has the structure

#sqrt(a+bsqrt(3))=csqrt(3)+d# so squaring both sides

#a+bsqrt(3)=3 c^2+ 2 sqrt[3] c d + d^2# pairing terms

#{(a - 3 c^2 - d^2 = 0),(b - 2 c d=0):}#

Solving for #c,d# we have

#c = pmsqrt[a pm sqrt[a^2 - 3 b^2]]/sqrt[6]#
#d=pm(sqrt[3/2] b)/sqrt[a - sqrt[a^2 pm 3 b^2]]#

If #a=4, b=2# we have the possibilities

#((c= -1/sqrt[3], d= -sqrt[3]),(c= 1/sqrt[3], d= sqrt[3]),(c= -1, d = -1),(c= 1, d = 1))#

Note that:

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

#color(white)((sqrt(3)+1)^2) = 3+2 sqrt(3)+1#

#color(white)((sqrt(3)+1)^2) = 4+2sqrt(3)#

Since #sqrt(3)+1 > 0#, we can take the positive square root of both ends to get:

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