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King
 4 years ago
its a sqrt question so i shall post it as a reply
King
 4 years ago
its a sqrt question so i shall post it as a reply

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^2x = 1 => x(x1) = 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i feel u can solve it

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see x is having number of sqrt that is upto infinity so let us take \[x = \sqrt{1+x}\] square both sides u will get \[x^2 = 1 + x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx help in my next question pls

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