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King

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

  • 2 years ago
  • 2 years ago

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

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

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

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

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

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

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

    • 2 years ago
  8. sheg
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    and x = 1

    • 2 years ago
  9. King
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    so which option

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

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

    • 2 years ago
  12. King
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    ok thnx

    • 2 years ago
  13. sheg
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    welcome :)

    • 2 years ago
  14. agdgdgdgwngo
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    hmm good question

    • 2 years ago
  15. agdgdgdgwngo
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    yeah sheg is right

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

    • 2 years ago
  17. agdgdgdgwngo
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    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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