Question

plz answer
options
3
5
6
2​

Answer 1

Answer:

3

Step-by-step explanation:

Let

$y = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...... + \infty } } }$

$= > y = \sqrt{6 + y}$

$= > {y}^{2} = 6 + y$

$= > {y}^{2} - y - 6 = 0$

$= > {y}^{2} - 3y + 2y - 6 = 0$

$= > y(y - 3) + 2(y - 3) = 0$

$= > (y - 3)(y + 2) = 0$

$either \: \: y = 3 \: \: or \: \: y = - 2$

since, square root is a positive quantity,

so we have y=3

Answer 2

Answer:

infenty I think sooooo