If $a = \frac{\sqrt{3}}{2}$, then $\sqrt{1+a} + \sqrt{1-a} = $ $x$

Concept & Formula Instead of plugging in the complex fraction directly under nested radicals, it is much easier to square the entire target expression. This takes advantage of the $(x+y)^2$ identity to eliminate the outer square roots and simplify the inner terms via the difference of squares. $$ (x+y)^2 = x^2 + y^2 + 2xy $$ $$ (1+a)(1-a) = 1 - a^2 $$ Step-by-Step Solution * **Given:** $a = \frac{\sqrt{3}}{2}$ and we need to evaluate $y = \sqrt{1+a} + \sqrt{1-a}$ * **Calculation:** Square the entire expression we want to evaluate: $y^2 = (\sqrt{1+a} + \sqrt{1-a})^2$ $y^2 = (\sqrt{1+a})^2 + (\sqrt{1-a})^2 + 2(\sqrt{1+a})(\sqrt{1-a})$ $y^2 = (1+a) + (1-a) + 2\sqrt{(1+a)(1-a)}$ * Simplify the terms: $y^2 = 2 + 2\sqrt{1 - a^2}$ * Now, substitute the given value of $a$: Since $a = \frac{\sqrt{3}}{2}$, then $a^2 = \frac{3}{4}$ * Plug $a^2$ into our simplified equation: $y^2 = 2 + 2\sqrt{1 - \frac{3}{4}}$ $y^2 = 2 + 2\sqrt{\frac{1}{4}}$ $y^2 = 2 + 2(\frac{1}{2})$ $y^2 = 2 + 1$ $y^2 = 3$ * Solve for $y$ (the original expression): $y = \sqrt{3}$ (Taking the positive root since the sum of positive square roots must be positive). Exam Strategy & Shortcut Whenever you see a pattern like $\sqrt{1+a} \pm \sqrt{1-a}$, immediately square the expression. It elegantly collapses the $a$ terms and creates a simple $1-a^2$ under the remaining radical. This is a standard competitive exam trick that bypasses nested radical denesting algorithms entirely. Common Pitfall Attempting to evaluate $\sqrt{1 + \frac{\sqrt{3}}{2}}$ directly requires finding a perfect square in the form of $(x+y)^2$ hidden under the root (e.g., multiplying top and bottom by 2 to get $\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}$). While mathematically possible, it is incredibly time-consuming and prone to arithmetic errors. Final Answer **Therefore, the correct answer is \sqrt{3}.**