Discussion

What is the value of \(\sqrt{6+\sqrt{6+\sqrt{6+...}}}\) ?

Answer: B

Let \(x = \sqrt{6+\sqrt{6+\sqrt{6+...}}}\).

Then, we can write \(x = \sqrt{6+x}\).

Squaring both sides, we get \(x^2 = 6+x\).

\(x^2 - x - 6 = 0\).

Factoring the quadratic equation: \((x-3)(x+2) = 0\).

Since the value inside the square root is positive, x must be positive. So, x = 3.

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