If #(x+sqrt(x^2+1))(y+sqrt(y^2+1))=1# what is #x+y#?

2 Answers
Sep 3, 2016

Answer:

#x+y = 0#

Explanation:

Let #y = -x#

Then:

#(x+sqrt(x^2+1))(y+sqrt(y^2+1)) = (sqrt(x^2+1)+x)(sqrt(x^2+1)-x)#

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

#color(white)((x+sqrt(x^2+1))(y+sqrt(y^2+1))) = 1#

So #x+y = 0# always results in #(x+sqrt(x^2+1))(y+sqrt(y^2+1)) = 1#

To see that this is the only possible value of #x+y#, note that the function #f(x) = x + sqrt(x^2+1)# is strictly monotonic increasing.

So if #x_1+sqrt(x_1^2+1) = x_2+sqrt(x_2^2+1)# then #x_1 = x_2#

Sep 3, 2016

Answer:

#x+y=0#

Explanation:

If #x+sqrt(x^2+1) = delta# then #y+sqrt(y^2+1) = 1/delta#

Solving

#(x-delta)^2=x^2+1# and

#(y-1/delta)^2=y^2+1#

we obtain

#x = (delta^2-1)/(2delta)# and
#y =- (delta^2-1)/(2delta)#

then

#x+y=0#