Question

$proved \: \: that \\ \: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + ....... \alpha } } } = 3$
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Answer 1

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Answer 2

$\huge\mathfrak\red{Answer}$

$\\ let \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \alpha } } } } = x \\ squaring \: on \: both \: side \\ \\6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \alpha } } } = {x}^{2} \\ \\ we \: know \: that \\ \sqrt{6 + \sqrt{6 + \sqrt{6.... \alpha } } } = x \\ so \\ 6 + x = {x}^{2} \\ \\ {x}^{2} - x - 6 = 0 \\ {x}^{2} - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ (x - 3)(x + 2) = 0 \\ we \: get \: x = - 2 \\ x = 3 \\ \\ so \: \\ \sqrt{6 + \sqrt{6 + \sqrt{6.... \alpha } } } = - 2 \\ which \: is \: not \: possible \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6...... \alpha } } } = 3$