Question

please give ans,asfast as possible

Answer 1

Answer:

(C).

Step-by-step explanation:

= > $x + \sqrt{x + \sqrt{x + \sqrt{ {x}^{} ...} } } = 2$ ... (1)

= > $\sqrt{x + \sqrt{x + \sqrt{ {x}^{} ...} } } = 2-x$

Square on both sides,

= > $(\sqrt{x + \sqrt{x + \sqrt{ {x}^{} ...} } })^2 = (2-x)^2$

= > $x + \sqrt{x + \sqrt{x + \sqrt{ {x}^{} ...} } } = (2-x)^2$

As given, LHS, is 2, from (1):

= > 2 = ( 2 - x )^2

= > ± √2 = 2 - x

= > x = 2 ± √2

Option C