Question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + .......} } }$
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Answer 1

$Answer \: \: \\ \\ Given \: \: Question \: \: Is \: \: \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } } } \\ \\ let \: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } = x \\ \\ \sqrt{6 + \sqrt{ 6 + \sqrt{6 + ... \infty } } } = x \\ \\ \sqrt{6 + x} = x \\ Squaring \: \: on \: both \: sides \: we \: have \\ \\ ( \sqrt{6 + x} ) {}^{2} = x {}^{2} \\ \\ 6 + x = x {}^{2} \\ \\ x {}^{2} - x - 6 = 0 \\ \\ x {}^{2} - 3x + 2x - 6 = 0 \\ \\ x(x - 3) + 2(x - 3) = 0 \\ \\ (x - 3) = 0 \: \: \: \: \: or \: \: \: \: (x + 2) = 0 \\ \\ x = 3 \: \: \: or \: \: \: x = - 2 \\ \\ x = - 2 will be rejected \: becoz \: \: domain \: or \: input \: of \: \\ square \: root \: function \: is \: set \: of \: real \: numbers \\ \\ so \: \: \: x = 3 \\ \\ therefore \: \: \: sum \: of \: this \: series \: is \: \: 3$