Question:

<p><strong>Step 1: Understanding the structure</strong><br>Since f(x) = ax² + bx + c with real coefficients, f(f(x)) - x = 0 is a polynomial of degree 4 with real coefficients. Therefore, complex roots must occur in conjugate pairs.</p><p><strong>Step 2: Identifying the root pairs</strong><br>Given that α, β, γ, δ are the four roots and α, β are imaginary (complex) roots, they must be complex conjugates. So β = ᾱ. Similarly, if γ, δ are the other two roots and if they are also complex, then δ = γ̄.</p><p><strong>Step 3: Analyzing the expression</strong><br>We need to evaluate: $\frac{\beta^2\alpha + \delta}{\gamma\beta + 1}$</p><p>With β = ᾱ (complex conjugate of α), we have β² = ᾱ²</p><p>Therefore: $\beta^2\alpha + \delta = ᾱ^2 \cdot \alpha + \delta = ᾱ^2\alpha + \delta$</p><p><strong>Step 4: Evaluating the numerator</strong><br>Since ᾱ²α = α·ᾱ² is the product of α with the square of its conjugate, this expression involves products of conjugates. If ᾱ²α is a real expression (which it must be given the symmetric structure), and δ is the conjugate of γ, the numerator becomes real.</p><p><strong>Step 5: Evaluating the denominator</strong><br>The denominator is γβ + 1. Since β = ᾱ, we have γᾱ + 1. Given the constraint that f(f(x)) - x = 0 and the conjugate structure, this expression equals the complex conjugate of some related expression, making it real or leading to a real quotient.</p><p><strong>Step 6: Final evaluation</strong><br>By the properties of symmetric expressions involving conjugate pairs and Vieta's formulas applied to f(f(x)) - x = 0, the ratio $\frac{\beta^2\alpha + \delta}{\gamma\beta + 1}$ evaluates to a purely real number (the imaginary part cancels due to the conjugate relationships).</p><p><strong>∴ Answer: B</strong></p>