<p>71. If the equation \(ax^2 + bx + c = x\) has no real roots, then the equation \(a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c = x\) will have</p>
Step-by-Step Solution
Key Concept: If ax² + bx + c = x has no real roots, then ax² + (b-1)x + c = 0 has no real roots. Substituting y = ax² + bx + c transforms the second equation into ay² + by + c = x, which becomes ay² + (b-1)y + c = 0 in terms of y—this has the same form as the original, hence no real roots for y. Since y = ax² + bx + c has no real values, x cannot be real.
<p><strong>Step 1:</strong> The given condition states ax² + bx + c = x has no real roots.</p><p>This means ax² + (b-1)x + c = 0 has discriminant Δ₁ = (b-1)² - 4ac < 0.</p><p><strong>Step 2:</strong> Let y = ax² + bx + c. The second equation becomes a(y)² + b(y) + c = x, or equivalently ay² + (b-1)y + c = 0 (rewriting as ay² + by + c = y).</p><p><strong>Step 3:</strong> This quadratic in y has discriminant Δ₂ = (b-1)² - 4ac < 0 (same as Δ₁).</p><p><strong>Step 4:</strong> However, the critical insight: for any real x, y = ax² + bx + c is a real number. But since ay² + (b-1)y + c = 0 has no real solutions for y, there is no real value of y that satisfies the equation.</p><p><strong>Step 5:</strong> Therefore, there is no real x such that y = ax² + bx + c satisfies the transformed equation simultaneously.</p><p>∴ Answer: <strong>No real roots</strong> (or 0 real roots)</p>
Correct Answer: 2