If #x=(sqrt(7)+sqrt(3))/(sqrt(7)-sqrt(3))# and #xy=1# then show that # ?

If #x=(sqrt(7)+sqrt(3))/(sqrt(7)-sqrt(3))# and #xy=1# then show that #(x^2+xy+y^2)/(x^2-xy+y^2)=12/11#

1 Answer
George C.
Dec 9, 2017

See explanation...

Explanation:

There may be a simpler way, but here's a direct approach...

Given:

#x = (sqrt(7)+sqrt(3))/(sqrt(7)-sqrt(3))#

and:

#xy = 1#

Note that:

#y = 1/x = (sqrt(7)-sqrt(3))/(sqrt(7)+sqrt(3))#

Rationalising the denominators, we find:

#x = 1/4(sqrt(7)+sqrt(3))^2 = 1/4(10+2sqrt(21)) = 1/2(5+sqrt(21))#

#y = 1/4(sqrt(7)-sqrt(3))^2 = 1/4(10-2sqrt(21)) = 1/2(5-sqrt(21))#

So:

#x^2+xy+y^2 = 1/4(5+sqrt(21))^2+1+1/4(5-sqrt(21))^2#

#color(white)(x^2+xy+y^2) = 1/4(46+2sqrt(21))+1+1/4(46-2sqrt(21))#

#color(white)(x^2+xy+y^2) = 1/2(23+sqrt(21))+1+1/2(23-sqrt(21))#

#color(white)(x^2+xy+y^2) = 24#

#x^2-xy+y^2 = 1/4(5+sqrt(21))^2-1+1/4(5-sqrt(21))^2#

#color(white)(x^2-xy+y^2) = 1/4(46+2sqrt(21))-1+1/4(46-2sqrt(21))#

#color(white)(x^2-xy+y^2) = 1/2(23+sqrt(21))-1+1/2(23-sqrt(21))#

#color(white)(x^2-xy+y^2) = 22#

So:

#(x^2+xy+y^2)/(x^2-xy+y^2) = 24/22 = 12/11#

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