If x = #sqrt(6+sqrt(6+sqrt(6+...))#) and y= #sqrt(6-sqrt(6-sqrt(6-...)))# what is #x-y# ?

1 Answer

#x-y=1#

Explanation:

Given that

#x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}#

#x=\sqrt{6+x}#

#x^2=6+x#

#x^2-x-6=0#

#x^2-3x+2x-6=0#

#x(x-3)+2(x-3)=0#

#(x-3)(x+2)=0#

#x=-2, 3#

But, #x>0#, hence

#x=3#

Similarly,

#y=\sqrt{6-\sqrt{6-\sqrt{6-\ldots}}}#

#y=\sqrt{6-y}#

#y^2=6-y#

#y^2+y-6=0#

#y^2+3y-2y-6=0#

#y(y+3)-2(y+3)=0#

#(y+3)(y-2)=0#

#y=-3, 2#

But, #y>0#, hence

#y=2#

Now,

#x-y#

#=3-2#

#=1#