Question

say me the value for this question ​

Answer 1

Answer:

1 + √2

Step-by-step explanation:

$Given:x =\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{1+...}}}}$

$Assume,x = \sqrt{1+2\sqrt{1+2\sqrt{1+...}}}$

Now,

As the expression under square root extends infinitely then expression in brackets would equal to x itself, so we can write the given expression as:  

$=>x = \sqrt{1+2x}$

On squaring both sides, we get

$=>x^2 =1 + 2x$

$=>x^2 - 2x - 1 = 0$

Here, a = 1, b = -2, c = -1.

D = b² - 4ac

   = (-2)² + 4

   = 8

The solutions are:

x = -b ± √D/2a

  = -(-2) ± √8/2

  = 1 ± √2

  = 1 + √2, 1 - √2

Since x cannot be negative, x = 1 + √2.

So, the final answer is 1 + √2.

Hope it helps!

Answer 2

√1+ 2x = x

1 + 2x = x^2

x^2 -2x -1 = 0

x = 2 +-√4 +4)/ 2

= 2+-√8)/2

= 2+- 2√2)/2

= 1 +√2