Question

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

Answer 1

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + .... \infty } } } } = x(let) \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{ 6 + \sqrt{6}....... \infty } } } = x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x$
√{6 + x} = x
6 + x = x²
x² - x - 6 = 0
x² - 3x + 2x - 6 = 0
x(x - 3) + 2(x - 3) = 0
(x + 2)(x - 3) = 0
x = - 2, 3
because x ≠ -2 { square root can't be negative.}

hance, the answer is x = 3

Answer 2

The ans is 3. because square roots won't be negative.

:-)hope it helps u.