Question

$\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 = } } } } }$
find???​

Answer 1

Step-by-step explanation: Hey mate here is detail explaination of your question plz mark as brainliest

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

Let it be equation (1)

Assume equation (1) as 'x'

So we get,

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

On squaring both the sides we get,

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

And equation (1) as 'x'and again we got the same situation so we can write it as,

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

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

On factorising this using the method of splitting the middle term we get,

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

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

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

So here we get two values of 'x' i. e

2 and -1

On taking out square root the value is always positive. So we conclude that,

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