If #(5-sqrt(x))^2 = y-20sqrt(2)# where #x, y# are integers, then what are #x# and #y# ?

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Answer 1 · 2015-06-03T23:08:38

Expand and equate the irrational parts to help simplify and find:

#x = 8# and #y = 33#

Expand the left hand side:

#(5 - sqrt(x))^2#

#= 5^2-(2xx5xxsqrt(x))+x#

#= 25 - 10sqrt(x)+x#

#=(x+25)-10 sqrt(x)#

In order to eliminate the irrational #-20 sqrt(2)# term with #x# and #y# being integers, we must have:

#-10 sqrt(x) = -20 sqrt(2)#

Divide both sides by #-10# to get:

#sqrt(x) = 2sqrt(2)#

So

#x = (2sqrt(2))^2 = 8#

#y = x+25 = 8+25 = 33#

#color(white)()#
Footnote

Why is it possible to equate the irrational parts like this?

Consider the set of all numbers of the form #a+b sqrt(2)# where #a# and #b# are rational numbers.

These representations are unique:

Suppose #a+b sqrt(2) = c+d sqrt(2)#

Then #a-c = (d-b)sqrt(2)#

So if #d != b# we would find #sqrt(2) = (a-c)/(d-b)# which is a rational number, but #sqrt(2)# is irrational.

So we must have #d=b# and hence #a=c#.