39no such x exists
atleast thats wat i am getting
10 Answers
62x = √4+√4-√4+......∞
y = √4-√4+√4+......∞
x=√4+y
y=√4-x
4-y2=x
x2-4=y
(x+y)(x-y)=x+y
so x-y=1
so we have
x2-4=x-1
x2-x-3=0
x=1±√132
ignore the -ve value [1]
62for the first 10 minutes i had no clue on how to go about solving this one..
Then i recalled a problem on similar lines
x+√y=7
and y+√x=11
My friend had given it to me when i was preparing for olympiads.. [1]
1hi nishanth
i am not having anything in my mind except that
we have to solve for :
x=(y-11)^2 and
y=(x-7)^2
we have to consider only the root that satisfies
y<11 and x<7
sorry if the answer is dumb