If x − y = 6 and xy = 40 for real numbers x and y, find the value of x² + y².

Introduction / Context:
This algebra problem uses symmetric expressions in two variables and standard identities relating the difference and product of two numbers to the sum of their squares. Instead of solving for x and y individually, which is possible but longer, we can use identities to find x² + y² directly. This technique is widely applicable in aptitude tests and helps to save time.

Given Data / Assumptions:
- x and y are real numbers.
- x − y = 6.
- xy = 40.
- We are asked to compute x² + y².

Concept / Approach:
The identity involving the square of a difference is:
(x − y)² = x² + y² − 2xy.
We know the left hand side and the product xy, so we can rearrange this identity to express x² + y² in terms of x − y and xy. Rearranging gives x² + y² = (x − y)² + 2xy. After computing these values, we can directly obtain the result without solving for x and y separately.

Step-by-Step Solution:
Step 1: Start with the identity (x − y)² = x² + y² − 2xy. Step 2: Rearrange this to express x² + y²: x² + y² = (x − y)² + 2xy. Step 3: Substitute the given values x − y = 6 and xy = 40 into this formula. Step 4: Compute (x − y)² = 6² = 36. Step 5: Compute 2xy = 2 * 40 = 80. Step 6: Add these two results: x² + y² = 36 + 80 = 116. Step 7: Therefore, x² + y² equals 116.
Verification / Alternative check:
We can find x and y explicitly to double check. Let x and y be roots of a quadratic equation. We have x − y = 6 and xy = 40. Let s = x + y. Then (x − y)² = x² + y² − 2xy and (x + y)² = x² + y² + 2xy. Adding these gives (x − y)² + (x + y)² = 2(x² + y²). Using x − y = 6 and xy = 40, we already know x² + y² = 116, so (x + y)² = x² + y² + 2xy = 116 + 80 = 196, giving x + y = 14 or x + y = −14. Solving x − y = 6 and x + y = 14 yields x = 10, y = 4, and indeed x² + y² = 100 + 16 = 116, which confirms our earlier result.

Why Other Options Are Wrong:
Option b, 80, corresponds only to 2xy and ignores the contribution from (x − y)².
Option c, 89, and option e, 100, do not satisfy the identity when you plug back in the known values of x − y and xy.
Option d, 146, is larger than the correct sum and does not match any consistent pair of x and y that obey the given conditions.

Common Pitfalls:
Students sometimes attempt to solve for x and y first, which is longer and presents more opportunities for arithmetic mistakes. Another frequent error is using the wrong identity, such as mistaking (x − y)² = x² − y² instead of the correct x² + y² − 2xy. Also, forgetting the sign of 2xy in the rearranged identity can lead to an incorrect sum. Working symbolically with the identities and substituting carefully helps avoid these issues.

Final Answer:
The value of x² + y² is 116.