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



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King
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\[x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}...\]

sheg
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\[x^2x = 1 => x(x1) = 1\]

sheg
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now i feel u can solve it

King
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yeah but how u got the x2 thing equation

King
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ok btw options are x=1, 0<x<1, 1<x<2, x is infinite

sheg
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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
\[x^2 = 1 + x\]

sheg
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sorry x1 = 1
x = 2

sheg
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and x = 1

King
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so which option

King
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1<x<2?

sheg
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yes i too feel so

King
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ok thnx

sheg
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welcome :)

agdgdgdgwngo
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hmm good question

agdgdgdgwngo
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yeah sheg is right

King
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thnx help in my next question pls

agdgdgdgwngo
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it converges to 1.75487766625 so the value of x should be in the range 1 < x < 2