## King 4 years ago its a sqrt question so i shall post it as a reply

1. anonymous

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

2. anonymous

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

3. anonymous

now i feel u can solve it

4. anonymous

yeah but how u got the x2 thing equation

5. anonymous

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

6. anonymous

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. anonymous

sorry x-1 = 1 x = 2

8. anonymous

and x = 1

9. anonymous

so which option

10. anonymous

1<x<2?

11. anonymous

yes i too feel so

12. anonymous

ok thnx

13. anonymous

welcome :)

14. anonymous

hmm good question

15. anonymous

yeah sheg is right

16. anonymous

thnx help in my next question pls

17. anonymous

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

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