$\sqrt{17956} + \sqrt{24025} = x$

### Concept & Strategy This requires finding the square roots of two separate large numbers and summing them. Use the unit digit technique combined with base range estimation for quick extractions. Base approximation logic: $$ \text{Estimate } a^2 \text{ and } b^2 \text{ to bound the root.} $$ ### Step-by-Step Solution - **Part 1: Find $\sqrt{24025}$** - The number ends in $25$, which means its square root must end in $5$. - Remove the last two digits to get $240$. - Find two consecutive integers whose product is close to but less than or equal to $240$. $15 \times 16 = 240$. - This perfectly matches the pattern for numbers ending in $5$. The square root is $155$. - **Part 2: Find $\sqrt{17956}$** - The number ends in $6$, so its root ends in $4$ or $6$. - Approximate the base: $130^2 = 16900$ and $140^2 = 19600$. - $17956$ is between them. Let's test $134$ and $136$. - $135^2 = 13 \times 14 \text{ appended with } 25 = 18225$. - Since $17956 < 18225$, the root must be less than $135$. Thus, it is $134$. - **Part 3: Addition** - Sum the two roots: $134 + 155 = 289$. ### Exam Strategy & Shortcut Use the unit digit of the final sum. The first root ends in $4$. The second root ends in $5$. The sum of their unit digits is $4 + 5 = 9$. Looking at the options, only $19$ and $289$ end in $9$. Since both roots are clearly greater than $100$, the sum must be over $200$, making $289$ the only logical choice. ### Common Pitfall Miscalculating the square root of the first term by choosing the upper boundary ($136$) instead of the lower boundary ($134$). Comparing against the midpoint square ($135^2$) is the most reliable way to prevent this error. ### Final Answer Therefore, the correct answer is 289.