## King 3 years ago its a sqrt question so i shall post it as a reply

1. King

$x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}...$

2. sheg

$x^2-x = 1 => x(x-1) = 1$

3. sheg

now i feel u can solve it

4. King

yeah but how u got the x2 thing equation

5. King

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

6. sheg

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

sorry x-1 = 1 x = 2

8. sheg

and x = 1

9. King

so which option

10. King

1<x<2?

11. sheg

yes i too feel so

12. King

ok thnx

13. sheg

welcome :)

14. agdgdgdgwngo

hmm good question

15. agdgdgdgwngo

yeah sheg is right

16. King

thnx help in my next question pls

17. agdgdgdgwngo

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