Question

√12+√12+√12+√12+√12....... infinity​

Answer 1

$let \: \: \sqrt{12 + \sqrt{12 + \sqrt{12 + ...............} } } = x \\ squaring \: \: both \: \: sides \\ = > 12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + .............} } } = x^{2} \\ = > 12 + x = {x}^{2} \\ = > x^{2} - x - 12 = 0 \\ = > {x}^{2} - 4x + 3x - 12 = 0 \\ = > (x - 4)( x + 3) = 0$

Thus, x = 4 or x = -3

But, x = -3 is not possible. So, x = 4.

Hope, it'll help you.....