Question

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

Answer 1

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

Answer 2

\[x^2-x = 1 => x(x-1) = 1\]

Answer 3

now i feel u can solve it

Answer 4

yeah but how u got the x2 thing equation

Answer 5

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

Answer 6

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
\[x^2 = 1 + x\]

Answer 7

sorry x-1 = 1
x = 2

Answer 8

and x = 1

Answer 9

so which option

Answer 10

1<x<2?

Answer 11

yes i too feel so

Answer 12

ok thnx

Answer 13

welcome :)

Answer 14

hmm good question

Answer 15

yeah sheg is right

Answer 16

thnx help in my next question pls

Answer 17

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