Beauty of Maths

So here is the solution..

$\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}} - 1 = x$ .

Let $y$ = $\sqrt{2\sqrt{2\sqrt{2...}}}$

Rewriting the equation we obtain,

$y - 1$ = $x$

Squaring both sides,

$(y-1)^{2} = x^{2}$

Expanding the brackets of the Left Hand Side we obtain,

$y^{2} - 2{y} + 1 = x^{2}$

Calculating for $y^{2}$ we get that it is $( \sqrt{2\sqrt{2\sqrt{2...}}} )^{2}$

Or simply,

$2 \times \sqrt{2\sqrt{2\sqrt{2...}}}$

Calculating for $2y$ we attain that it is $2\sqrt{2\sqrt{2\sqrt{2...}}}$

We notice here that $y^{2} = 2y$ , and so the equation reduces to $x^{2} - 1 = 0$

Hence by Difference Of Two Squares, $(x-1)(x+1) = 0$

However, x = -1 is not a valid solution (extraneous root)

And the solution is: $\boxed{x=1}$

√[2√[2[√2.... = (x + 1) square both sides

2(x + 1) = x² + 2x + 1

2x + 2 = x² + 2x + 1

x² = 1, x = ± 1

2 (x + 1) = (x + 1)^2

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

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

=> x = 1 as ? - 1 cannot be 0 - 1.

When we take square root of 2 we get 1,41,mutiply by 2 we get2.82,again squareroot we get 1.67 multiply by 2 and srareroot we get finally 4 whose squareroot is2 then take minus 1 so X =1 Ans K.K.GARG,India