What is the value of?

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

1 Answer
George C.
Feb 24, 2018

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

Explanation:

Note that:

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

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

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

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

#color(white)(sqrt(2)(sqrt(2+sqrt(3))+sqrt(2-sqrt(3)))) = 2sqrt(3)#

Dividing both ends by #sqrt(2)# we find:

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

Footnote

Why multiply by #sqrt(2)# ?

When faced with an expression like #sqrt(2+sqrt(3))#, it would be simple for us if the #2+sqrt(3)# was recognisable as the square of something of the form #a+bsqrt(3)#.

However, if we square #a+bsqrt(3)# then we get:

#(a^2+3b^2)+2ab sqrt(3)#

Without getting into trying to solve for #a# and #b#, we can notice immediately that the coefficient of #sqrt(3)# is a multiple of #2#.

What if we multiply the radicand by #2# first?

Then we get:

#4+2sqrt(3)#

which is recognisable as:

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

In order to multiply the radicand by #2# we need to multiply #sqrt(2+sqrt(3))# by #sqrt(2)#.

So that's why.

Apologies for the minor but gratuitous rabbit-pulling.

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