$\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}$ is equal to

### Concept & Logic This evaluates an infinite repeating nested square root. When the repeating number can be expressed as the product of two consecutive integers, a rapid logical deduction can be made instead of full algebraic calculation. The core substitution logic remains: $$ y = \sqrt{a + y} $$ ### Step-by-Step Solution * Let the entire expression be $x$. * $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}$ * Because the nested roots continue infinitely, the part under the first square root after "$2 +$" is exactly equal to our original definition of $x$. * Replace the infinite tail with $x$: * $x = \sqrt{2 + x}$ * Square both sides of the equation: * $x^2 = 2 + x$ * Bring all terms to one side to form a quadratic equation: * $x^2 - x - 2 = 0$ * Factor the quadratic equation. We need two numbers that multiply to -2 and add to -1. Those numbers are -2 and 1. * $(x - 2)(x + 1) = 0$ * This gives two possible solutions: $x = 2$ or $x = -1$. * Since a square root function (and a sum of positive terms) must yield a positive result, $x$ cannot be negative. * We discard -1. Thus, $x = 2$. ### Exam Strategy & Shortcut Whenever you face an infinite series in the form $\sqrt{a + \sqrt{a + \sqrt{a + \cdots}}}$, immediately check if $a$ can be factored into two consecutive integers ($n \times (n+1)$). If it is a "+" series, the answer is always the larger integer $(n+1)$. If it is a "-" series, the answer is the smaller integer $(n)$. Here, $2 = 1 \times 2$. Since it's a "+" series, the answer is the larger factor, which is 2. Zero calculations required. ### Common Pitfall The most frequent error is choosing the smaller factor (1) instead of the larger factor (2). Remember: addition sequences yield the larger consecutive factor, while subtraction sequences yield the smaller one. ### Final Answer **Therefore, the correct answer is 2.**