Question

infinit times.. ? is equal to? send answer in you notebook by writing plz dosto emergency

Answer 1

[0010001111]... Hello User... [1001010101]
Here's your answer...

$let \: \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } = x \\ then \: x = \sqrt{6 + x}$
Because if u put one more root 6 in infinity, it won't cause a change in value.
$x = \sqrt{6 + x} \\ {x}^{2} = 6 + x \\ {x}^{2} - x - 6 = 0 \\ (x - 3)(x + 2) = 0 \\ x = 3 \: or \: - 2$
But if the value -2 is put in the first equation, then...
$- 2 = \sqrt{6 - 2} \\ - 2 = \sqrt{4}$
When we take roots of numbers, we take them in absolute values. So -2 cannot be a solution to the equation.
The only solution is 3.

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