How can we attempt to simplify expressions of the form #sqrt(p+qsqrt(r))# where #p, q, r# are rational?

1 Answer
George C.
May 19, 2017

If #q > 0# and #p^2-q^2r# is a perfect square #s^2# then:

#sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2#

Explanation:

Consider the quartic:

#(x-sqrt(p+qsqrt(r)))(x+sqrt(p+qsqrt(r)))(x-sqrt(p-qsqrt(r)))(x+sqrt(p-qsqrt(r)))#

#=(x^2-(p+qsqrt(r)))(x^2-(p-qsqrt(r)))#

#=((x^2-p)-qsqrt(r))((x^2-p)+qsqrt(r)))#

#=(x^2-p)^2-q^2r#

#=x^4-2px^2+(p^2-q^2r)#

Suppose #p^2-q^2r = s^2# for some positive rational number #s#.

Since this quartic has no terms of odd degree it can alternatively be factored as:

#=(x^2-kx+s)(x^2+kx+s)#

#=x^4+(2s-k^2)x^2+s^2#

Equating coefficients, we have:

#-2p = 2s-k^2#

and hence:

#k^2 = 2p+2s#

By the quadratic formula, the zeros of the quadratic factors are:

#(+-k+-sqrt(k^2-4s))/2 = +-sqrt(2p+2s)/2+-sqrt(2p-2s)/2#

where all combinations of signs are allowed.

In particular, if #q > 0# then:

#sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2#

If #q < 0# then:

#sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2-sqrt(2p-2s)/2#

#color(white)()#
Example

Simplify:

#sqrt(5+2sqrt(6))#

#{ (p=5), (q=2), (r=6), (s=sqrt(p^2-q^2r)=sqrt(5^2-2^2*6)=1) :}#

So:

#sqrt(5+2sqrt(6)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2#

#color(white)(sqrt(5+2sqrt(6))) = sqrt(12)/2+sqrt(8)/2#

#color(white)(sqrt(5+2sqrt(6))) = sqrt(3)+sqrt(2)#

Related questions
See all questions in Applications of Quadratic Functions
Impact of this question
4669 views around the world
You can reuse this answer
Creative Commons License