What is the value of \(\sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}\)?
3
Let \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}\). The expression repeats inside, so:
\(x = \sqrt{6 + x}\)
Squaring both sides: \(x^2 = 6 + x \;\Rightarrow\; x^2 - x - 6 = 0\).
Factorising: \((x - 3)(x + 2) = 0 \;\Rightarrow\; x = 3\ \text{or}\ x = -2\).
Since x is a square root, x must be non-negative. Hence \(x = 3\).
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