If x and y are non-zero real numbers, then x² + xy + y²
2018
If x and y are non-zero real numbers, then x² + xy + y²
- A.
is always negative
- B.
takes the value zero for some x and y
- C.
is always positive
- D.
takes both positive and negative values
Attempted by 3 students.
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Correct answer: C
Concept: For a quadratic expression in two variables like ax² + bxy + cy², sign-definiteness (always positive, always negative, or mixed) can be tested by completing the square — rewriting it as a sum of squared terms. If the rewritten form is a sum where every term is non-negative and at least one term is strictly positive whenever the variables are non-zero, the original expression is always positive.
Treat x² + xy + y² as a quadratic in x and complete the square: x² + xy + y² = (x + y/2)² + (3/4)y².
The squared term (x + y/2)² is a square of a real number, so it is always ≥ 0, for any real x and y.
Since y is given to be a non-zero real number, y² > 0, so (3/4)y² is strictly positive.
Adding a term that is ≥ 0 to a term that is strictly > 0 gives a strictly positive sum, so x² + xy + y² > 0 for every non-zero real x and y — the sign does not depend on whether x or y is positive or negative.
x = 1, y = 1: 1² + (1)(1) + 1² = 1 + 1 + 1 = 3 > 0.
x = 1, y = -1 (opposite signs): 1² + (1)(-1) + (-1)² = 1 - 1 + 1 = 1 > 0.
x = -2, y = 4 (chosen so x + y/2 = 0, the boundary where the squared term vanishes): (-2)² + (-2)(4) + 4² = 4 - 8 + 16 = 12 > 0, confirming positivity holds even at this boundary case.
So x² + xy + y² is always positive whenever x and y are non-zero real numbers.