Question

30.
$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2............. \infty = } } } }$

1) 1
22
3) 4
4)8​

Answer 1

Answer:

let x= given expression. squareing both sides.

x^2=2+x

x^2-x-2=0

(x-2)(x+1)=0

x=2 or -1

x=-1 is impossible.since x is positive.

There fore x=2

Answer 2

Step-by-step explanation:

GIVEN

$x = \sqrt{2 + \sqrt{2} + \sqrt{2} + ...... \infty }$

see in picture

repeated red area is exactly

same as x

☆ then we can substitute X into the red area ,

Now we have a simple equation

⬅️$</strong><strong>X</strong><strong> </strong><strong> = \sqrt{2 + </strong><strong>X</strong><strong>}$

squaring both side of the equation

${X}^{2} = ( \sqrt{2 + X} ) {}^{2}$

${x}^{2} = 2 + x \: \: ..............(i)$

${x}^{2} - x - 2$

${x}^{2} - 2x + x + 2 = 0$

$x(x - 2) + 1(x - 2) = 0$

$(x - 2)(x + 1) = 0$

x = 2 , -1

here we use 2 for x because x^2 can never be negative.

NOW

value of x put in equation (i)

${2}^{2} = 2 + 2$

$4 = 4$

so , the equation is true

then solution of this question is x = 2

hence option 2 is the right answer ✔