Math, asked by madhavanaalvala4769, 10 months ago

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

1) 1
22
3) 4
4)8​

Answers

Answered by sprao534
2

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

Answered by Anonymous
13

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

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