It says that converges to k not that is the value.
\(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Squaring both sides is the best approach for such questions
\(x^2 = 6 + \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)
Now since we know this goes on infinitely, we can say that the part after the 6 in the right hand side of the equation is equal to the original
xRewrite the equation as
\(x^2 = 6 + x\)
\(x^2 - x - 6 = 0\)
\(x^2 + 2x - 3x - 6 = 0\)
\((x+2)(x-3) = 0\)
x = -2 or 3
Since x is a square root expression, it cannot be negative so
\(x = 3\) is the only solution
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