Question

17. If x=2 + root2
root2+root2+root2+root2+
and so on
then x is equal to
​

Answer 1

$x = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}. \: . \: . } } } \\ let \: y = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}. \: . \: . } } } \\ then \: y \: = \sqrt{2 + y} \\ {y}^{2} = 2 + y \\ {y}^{2} - y - 2 = 0 \\ {y}^{2} - 2y - 1y - 2 = 0 \\ y(y - 2) - 1(y - 2) = 0 \\ (y - 1)(y - 2) = 0 \\ y = 1 \: or \: 2 \\ \\ when \: y = 1 \: \\ x = 2 + 1 = 3 \\ \\ when \: y = 2 \\ x = 2 + 2 = 4$

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