Question

The value of
$\sqrt{6 + \sqrt{6 + \sqrt{6 + ...} } }$
is ???

Most correct and fastest answer = Brainliest Answer ☺ ​

Answer 1

Answer:

x = -2, 3

Step-by-step explanation:

Squaring both sides you'll get,

x^2 = 6+x

x^2 - x - 6 = 0

x^2 +2x - 3x = 0

Taking common,

x(x+2) - 3(x+2) = 0

(x+2)(x-3) = 0

x = -2,3

Answer 2

$[tex]let \: y \: = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...} } } \\ on \: squiring \: both \: side \: we \: get \\ {y}^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ...} } } \\ {y}^{2} - y - 6 = 0\\y^2 - 2y + 1/4 - 1/4 - 6 = 0\\(y - 1/2)^2 - (5/2)^2 = 0\\( y - 3)(y - 2) = 0 \\y = 3 \: and \: y = 2</p><p></p><p></p><p>$