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King

  • 4 years ago

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

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  1. King
    • 4 years ago
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    \[x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}...\]

  2. sheg
    • 4 years ago
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    \[x^2-x = 1 => x(x-1) = 1\]

  3. sheg
    • 4 years ago
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    now i feel u can solve it

  4. King
    • 4 years ago
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    yeah but how u got the x2 thing equation

  5. King
    • 4 years ago
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    ok btw options are x=1, 0<x<1, 1<x<2, x is infinite

  6. sheg
    • 4 years ago
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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 \[x^2 = 1 + x\]

  7. sheg
    • 4 years ago
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    sorry x-1 = 1 x = 2

  8. sheg
    • 4 years ago
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    and x = 1

  9. King
    • 4 years ago
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    so which option

  10. King
    • 4 years ago
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    1<x<2?

  11. sheg
    • 4 years ago
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    yes i too feel so

  12. King
    • 4 years ago
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    ok thnx

  13. sheg
    • 4 years ago
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    welcome :)

  14. agdgdgdgwngo
    • 4 years ago
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    hmm good question

  15. agdgdgdgwngo
    • 4 years ago
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    yeah sheg is right

  16. King
    • 4 years ago
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    thnx help in my next question pls

  17. agdgdgdgwngo
    • 4 years ago
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    it converges to 1.75487766625 so the value of x should be in the range 1 < x < 2

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