Nested Radicals

$\sqrt{ 12 + \sqrt{12 + \sqrt{ 12 + \cdots}}} =x$ $\Rightarrow \sqrt{12 + x} = x$ Squaring both sides , $12 + x = x^2$ $x^2 - x - 12 =0$ $(x + 3)(x - 4) = 0$ $\Rightarrow x + 3 = 0 \Rightarrow \boxed {x = -3}$ $\Rightarrow x - 4 = 0 \Rightarrow \boxed {x = 4}$ Now , 4 is the answer
And it is in the options also......

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Isn't it easy??

The root of the sum (12 + root12) is more than any answer other than 4, so because the actual answer is greater than the root of the sum (12 + root12), it has to be 4.

I found it in a way that is not rigorous at all.

First you consider the square root of 12 which is roughly 3.46 You add that to 12 to get 15.46, whose square root is almost 4. From that you can see from the choices of answers that 4 is the most likely as it is addition =).

Thinking that the positive root of 9 is 3. Also, the sum of a positive value inside a square root only increases its absolute value. These two analysis combined lead me to exclude numbers below 3 (1 and 2) as well as the root of 12. Therefore, only 4 could be the answer.

Rather than to calculate the exact answer for this we can choose the correct answer from given options easily. From the problem we can tell definitely the answer is more than √12 which is more than 3.464. From the given choices the answer satisfying the above statement is only 4.

Consider given expression as some 'x'.