Difficulty: Hard
Correct Answer: 1
Explanation:
Introduction / Context:This question tests algebraic transformation and completing the square in two variables. By moving all terms to one side and grouping x-terms and y-terms, we can rewrite the expression as a sum of squares. A sum of squares equals zero only when each square equals zero, letting us deduce exact values of x and y and then compute x - y.
Given Data / Assumptions:
Concept / Approach:Bring RHS to the left, then complete squares separately for x and y:(x^2 + 2x + 1) = (x + 1)^2 and (y^2 + 4y + 4) = (y + 2)^2. If the equation becomes (x + 1)^2 + (y + 2)^2 = 0, both squares must be 0, giving x and y uniquely.
Step-by-Step Solution: Start: x^2 + y^2 + 6x + 5 = 4(x - y) = 4x - 4y Move all terms to left: x^2 + y^2 + 6x + 5 - 4x + 4y = 0 Simplify: x^2 + y^2 + 2x + 4y + 5 = 0 Complete square in x: x^2 + 2x = (x + 1)^2 - 1 Complete square in y: y^2 + 4y = (y + 2)^2 - 4 Substitute: (x + 1)^2 - 1 + (y + 2)^2 - 4 + 5 = 0 Constants: -1 - 4 + 5 = 0, so (x + 1)^2 + (y + 2)^2 = 0 A sum of squares is 0 only if each square is 0: x + 1 = 0 => x = -1, and y + 2 = 0 => y = -2 Therefore x - y = (-1) - (-2) = 1
Verification / Alternative check:Substitute x = -1, y = -2 into original: LHS = 1 + 4 + 6*(-1) + 5 = 5 - 6 + 5 = 4. RHS = 4((-1) - (-2)) = 4*(1) = 4. Matches, so x - y = 1 is confirmed.
Why Other Options Are Wrong:0 and 2 come from mistakes in moving terms across the equal sign.-1 results from swapping y - x instead of x - y at the end.4 is typically from forgetting to complete the square correctly in both variables.
Common Pitfalls:Sign mistakes when expanding 4(x - y), and incorrect completion of squares (forgetting to add and subtract the same constant). Another error is assuming (x + 1)^2 + (y + 2)^2 = 0 has many solutions; it has exactly one real solution pair.
Final Answer:x - y = 1
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