Answer: B
Let \(x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}\)
\(\therefore x^2 = 12 + \sqrt{12 + \sqrt{12 + \cdots}} \)
That is, \(x^2 = 12 + x\)
So, \(x^2 - x - 12 = 0\)
Or, \(x^2 - 4x + 3x - 12 = 0\)
Or, \(x(x - 4) + 3(x - 4) = 0\)
Or, \((x - 4)(x + 3) = 0\)
\(\therefore\) Either \(x = 4\) or \(x = -3\)
Let \(x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}\)
\(\therefore x^2 = 12 + \sqrt{12 + \sqrt{12 + \cdots}} \)
That is, \(x^2 = 12 + x\)
So, \(x^2 - x - 12 = 0\)
Or, \(x^2 - 4x + 3x - 12 = 0\)
Or, \(x(x - 4) + 3(x - 4) = 0\)
Or, \((x - 4)(x + 3) = 0\)
\(\therefore\) Either \(x = 4\) or \(x = -3\)