Homogeneous quadratic in x and y (ratio form): If 2x^2 − 7xy + 3y^2 = 0 for nonzero x and y, determine the possible values of the ratio x : y.

Introduction / Context:
This problem asks for the ratio x : y that satisfies a homogeneous quadratic relation 2x^2 − 7xy + 3y^2 = 0. Because every term carries the same total degree, dividing through by y^2 (assuming y ≠ 0) reduces the equation to a quadratic in the single variable r = x / y. Solving that quadratic yields the admissible ratios, which must be presented as exact pairs x : y.

Given Data / Assumptions:

• Equation: 2x^2 − 7xy + 3y^2 = 0.
• x and y are real and not both zero; take y ≠ 0 for ratio.
• We need all possible real ratios x : y that satisfy the equation.

Concept / Approach:
Introduce r = x / y. Divide the equation by y^2 to obtain a standard quadratic in r. Solve using the quadratic formula and convert each r back into the ratio x : y = r : 1 (and, if r is a fraction p/q, express as p : q).

Step-by-Step Solution:

Verification / Alternative check:
Substitute r = 3: 2*(9) − 7*(3) + 3 = 18 − 21 + 3 = 0. Substitute r = 1/2: 2*(1/4) − 7*(1/2) + 3 = 0.5 − 3.5 + 3 = 0. Both satisfy the equation exactly.

Why Other Options Are Wrong:

• 3 : 2 or 2 : 3: These correspond to r = 3/2 or r = 2/3, neither satisfies 2r^2 − 7r + 3 = 0.
• 5 : 6: Gives r = 5/6, which does not solve the quadratic.

Common Pitfalls:
Forgetting to convert r values to simplified integer ratios, or losing a valid root by factoring incorrectly. Always check both quadratic roots and express as clean ratios.

Final Answer: