its a sqrt question so i shall post it as a reply

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its a sqrt question so i shall post it as a reply

Mathematics
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\[x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}...\]
\[x^2-x = 1 => x(x-1) = 1\]
now i feel u can solve it

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yeah but how u got the x2 thing equation
ok btw options are x=1, 0
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 \[x^2 = 1 + x\]
sorry x-1 = 1 x = 2
and x = 1
so which option
1
yes i too feel so
ok thnx
welcome :)
hmm good question
yeah sheg is right
thnx help in my next question pls
it converges to 1.75487766625 so the value of x should be in the range 1 < x < 2

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