Question

Solve this question No SPAMMING​

Answer 1

Let,

x = $\sf \sqrt{12 + \sqrt{12 + \sqrt{12 + }}}... \infty$

Now,

$</p><p> \sf \rightarrow \quad x = \sqrt{12 + \sqrt{12 + \sqrt{12+} } }... \infty \\ \\ \sf \rightarrow \quad x = \sqrt{12 + x} \\ \\ \sf \quad \because \: x = \sqrt{12 + \sqrt{12 + \sqrt{12 + } } } ... \infty \\ \\ \sf \rightarrow \quad x = \sqrt{12 + x} \\ \\ \sf \: square \: both \: sides \\ \\ \sf \rightarrow \quad {x}^{2} = 12 + x \\ \\ \sf \rightarrow \quad {x}^{2} - x - 12 = 0 \\ \\ \sf \rightarrow \quad {x}^{2} - 4x + 3x - 12 = 0 \\ \\ \sf \rightarrow \quad x(x - 4) + 3(x - 4) = 0 \\ \\ \sf \rightarrow \quad (x - 4)(x + 3) = 0 \\ \\ \sf \rightarrow \quad x = 4, \: - 3 </p><p>$

-3 is neglected because It is negative which is out of the range of the square root function.

So,

x = 4

Hence, Option (D) is correct.