The mean proportional between $(3 + \sqrt{2})$ and $(12 - \sqrt{32})$ is

### Concept & Formula The mean proportional between two terms $a$ and $b$ is $\sqrt{ab}$. When terms involve surds (radicals), simplify them first by pulling out common factors to reveal potential conjugate pairs. ### Step-by-Step Solution 1. Let $a = (3 + \sqrt{2})$ and $b = (12 - \sqrt{32})$. 2. Simplify the second term $b$. Note that $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$. So, $b = 12 - 4\sqrt{2}$. 3. Factor out the common $4$ from $b$: $b = 4(3 - \sqrt{2})$ 4. Set up the mean proportional equation $x = \sqrt{a \times b}$: $$ x = \sqrt{(3 + \sqrt{2}) \times 4(3 - \sqrt{2})} $$ 5. Rearrange to group the conjugate pairs together: $$ x = \sqrt{4 \times (3 + \sqrt{2})(3 - \sqrt{2})} $$ 6. Use the difference of squares identity $(A+B)(A-B) = A^2 - B^2$ on the conjugate pair: $(3 + \sqrt{2})(3 - \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7$ 7. Substitute this back into the square root equation: $$ x = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7} $$ ### Exam Strategy & Shortcut Recognize immediately that $12 - \sqrt{32}$ hides a multiple of the conjugate of $3 + \sqrt{2}$. Extracting $4$ transforms it to $4(3 - \sqrt{2})$. The product of conjugates $(a+b)(a-b)$ quickly yields an integer ($9-2=7$). Thus, $\sqrt{4 \times 7} = 2\sqrt{7}$ can be calculated almost entirely mentally. ### Common Pitfall Attempting to multiply $(3 + \sqrt{2})(12 - \sqrt{32})$ directly using FOIL without simplifying $\sqrt{32}$ first. This leads to a messy expression like $36 - 3\sqrt{32} + 12\sqrt{2} - \sqrt{64}$ which is highly prone to arithmetic errors. ### Final Answer Therefore, the correct answer is **$2\sqrt{7}$**.