The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is

### Concept & Formula This problem can be instantly solved by recognizing the standard algebraic identity for the square of a trinomial. $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) $$ ### Step-by-Step Solution * **Given:** The sum of the squares: $a^2 + b^2 + c^2 = 138$ The sum of the products two at a time: $ab + bc + ca = 131$ * **Calculation:** Substitute the given values into the algebraic identity. $$ (a + b + c)^2 = 138 + 2(131) $$ $$ (a + b + c)^2 = 138 + 262 $$ $$ (a + b + c)^2 = 400 $$ * **Deduction:** Taking the square root of both sides yields the sum of the numbers. $$ a + b + c = \sqrt{400} = 20 $$ ### Exam Strategy & Shortcut Memorize the structure $Sum^2 = SquaresSum + 2 \times ProductsSum$. In an exam, you should mentally calculate $138 + 2(131) = 400$, and immediately recognize that its square root is $20$. No writing is necessary. ### Common Pitfall A frequent mistake is forgetting to multiply the product sum by $2$ in the formula, leading to an incorrect sum of $138 + 131 = 269$, which doesn't yield a clean integer square root. ### Final Answer **Therefore, the correct answer is 20.**