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KingBest ResponseYou've already chosen the best response.0
\[x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}...\]
 2 years ago

shegBest ResponseYou've already chosen the best response.4
\[x^2x = 1 => x(x1) = 1\]
 2 years ago

KingBest ResponseYou've already chosen the best response.0
yeah but how u got the x2 thing equation
 2 years ago

KingBest ResponseYou've already chosen the best response.0
ok btw options are x=1, 0<x<1, 1<x<2, x is infinite
 2 years ago

shegBest ResponseYou've already chosen the best response.4
see x is having number of sqrt that is upto infinity so let us take \[x = \sqrt{1+x}\] square both sides u will get \[x^2 = 1 + x\]
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
yeah sheg is right
 2 years ago

KingBest ResponseYou've already chosen the best response.0
thnx help in my next question pls
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
it converges to 1.75487766625 so the value of x should be in the range 1 < x < 2
 2 years ago
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