If $a = \sqrt{3 + \sqrt{3 + \sqrt{3 + \cdots}}}$, then which of the following is true?

### Concept & Strategy This question asks you to establish the range of an infinite nested surd. Since the number 3 cannot be factored into two consecutive integers, we must use the general formula and then estimate the value of the resulting square root to find its bounds. ### Step-by-Step Solution * **Given:** $a = \sqrt{3 + \sqrt{3 + \sqrt{3 + \cdots}}}$ * We set up the equation using substitution: * $a = \sqrt{3 + a}$ * Squaring both sides: * $a^2 = 3 + a$ * Rearranging into a quadratic equation: * $a^2 - a - 3 = 0$ * We solve for $a$ using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with coefficients $1, -1, -3$: * $a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)}$ * $a = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}$ * Taking the positive root since $a$ must be positive: $a = \frac{1 + \sqrt{13}}{2}$. * Now, we must estimate $\sqrt{13}$ to find the range. * We know perfect squares: $\sqrt{9} = 3$ and $\sqrt{16} = 4$. * Therefore, $3 < \sqrt{13} < 4$. * Substitute these bounds into our expression for $a$: * Lower bound: $a > \frac{1 + 3}{2} \implies a > \frac{4}{2} \implies a > 2$ * Upper bound: $a < \frac{1 + 4}{2} \implies a < \frac{5}{2} \implies a < 2.5$ * Combining these gives $2 < a < 2.5$. * Looking at the options, the range that correctly encompasses this result is $2 < a < 3$. ### Exam Strategy & Shortcut You can instantly find the bounds by looking at the nearest numbers that CAN be factored into consecutive integers. We know that for $\sqrt{x + \sqrt{x + \cdots}}$, the answer is $n+1$ if $x = n(n+1)$. The nearest factorable number below 3 is 2 ($1 \times 2 \implies$ value is 2). The nearest factorable number above 3 is 6 ($2 \times 3 \implies$ value is 3). Since 3 lies between 2 and 6, the value of its infinite series must logically lie between 2 and 3. This bypasses the quadratic formula entirely. ### Common Pitfall Students often see the number 3 repeating and mistakenly choose $a = 3$, confusing the value of the term with the final evaluated sum of the series. Always bounds-test using known perfect consecutive products. ### Final Answer **Therefore, the correct answer is $2 < a < 3$.**