Let x and y be positive real numbers such that xy = 56 and x^2 + y^2 = 113. Using these conditions, find the exact value of the sum (x + y).

Introduction / Context:
This algebra question tests your understanding of how sums and products of two numbers relate to the squares of those numbers. Instead of solving directly for x and y, you are given xy and x^2 + y^2 and asked to find x + y. This kind of problem commonly appears in algebra and aptitude tests, often leading to a neat answer via a simple identity rather than lengthy calculations.

Given Data / Assumptions:

• x and y are positive real numbers.
• Their product is xy = 56.
• The sum of their squares is x^2 + y^2 = 113.
• We must find x + y.

Concept / Approach:
The main identity to use is: (x + y)^2 = x^2 + y^2 + 2xy. We are given both x^2 + y^2 and xy, so we can compute (x + y)^2 directly, then take the positive square root to find x + y. Because x and y are specified to be positive, x + y must also be positive, which resolves any ambiguity in taking square roots. This method is far faster than trying to form and solve a quadratic for x or y individually.

Step-by-Step Solution:
1) Start with the identity (x + y)^2 = x^2 + y^2 + 2xy. 2) Substitute the given values: x^2 + y^2 = 113 and xy = 56. 3) Compute 2xy = 2 * 56 = 112. 4) So (x + y)^2 = 113 + 112 = 225. 5) To find x + y, take the square root of both sides: x + y = ±√225. 6) Since x and y are positive, their sum x + y must also be positive, so x + y = √225 = 15.

Verification / Alternative check:
We can verify this result by explicitly solving for x and y. If x + y = 15 and xy = 56, then x and y are roots of the quadratic equation t^2 − 15t + 56 = 0. Factorizing gives (t − 7)(t − 8) = 0, so t = 7 or t = 8. Hence, {x, y} = {7, 8}. Now check the given conditions. The product xy = 7 * 8 = 56, and x^2 + y^2 = 7^2 + 8^2 = 49 + 64 = 113. Both constraints are satisfied, confirming that x + y = 7 + 8 = 15 is correct.

Why Other Options Are Wrong:
Option b (−15) would require x + y to be negative, which contradicts the fact that both x and y are positive. Options c (21), d (29), and e (13) lead to incorrect values of x^2 + y^2 or xy when plugged into the identity or used to form a quadratic equation. Only x + y = 15 satisfies both given conditions.consistently.

Common Pitfalls:
Some learners forget the identity for (x + y)^2 and attempt to solve for x and y directly, which is slower and more error prone. Others may mistakenly use x^2 + y^2 = (x + y)^2 − 2xy but then make sign errors when rearranging. Another frequent issue is forgetting to choose the positive square root for x + y, even though x and y are positive. Keeping the identity in mind and using the positivity condition avoids these mistakes.

Final Answer:
The sum of the two positive real numbers is 15.