Question

please help in following question
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Answer 1

Solution :-

So the question is

$\sqrt{6 + \sqrt{6 + \sqrt{6+ ...\infty}}}$

Now

Let us denote this expression with x

So

$x = \sqrt{6 + \sqrt{6 + \sqrt{6+ ...\infty}}}$

Now by squaring :-

$x^2 = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6+ ...\infty}}}$

Or

$x^2 = 6 + x$

So by solving this Quadratic equation.

$x^2 - x - 6 = 0$

$\implies x^2 + 2x - 3x - 6 = 0$

$\implies x(x + 2) - 3(x + 2) = 0$

$\implies (x+2)(x-3) = 0$

So x = -2 , 3

But sum of positive numbers can not be negative.

→ $\sqrt{6 + \sqrt{6 + \sqrt{6+ ...\infty}}} = 3$