Question

the value of√6+√6+√6+....................is please answer it's urgent​

Answer 1

$Let \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ......} } } } } = x$

$\: \: \: \: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ....} } } } = x \\ = > {\bigg( \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ....} } } }\bigg)}^{2} = {x}^{2} \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (squaring \: both \: sides) \\ = > 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ....} } } } = {x}^{2} \\ = > 6 + x = {x }^{2} \\ = > {x}^{2} - x - 6 = 0 \\ = > {x}^{2} - 3x + 2x - 6 = 0 \\ = > x(x - 3) + 2(x - 3) = 0 \\ = > (x + 2)(x - 3) = 0$

$either \: x + 2 = 0 \: \: \: \: \: \: \: \: \: \: \: || \: \: or \: x - 3 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: = > x = - 2 \: \: \: \: \: \: \: \: || = > \: \: \: \: \: x = 3$

$x \: cannot \: be \: negative \: as \: square \: root \: of \: any \: number \: cannot \: be \\ negative.$

$So, \: x = 3$

$\red{\boxed{\boxed{Answer: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ......} } } } } = 3}}}$