<p>Find the number of quadratic equations with real roots which remain unchanged even after squaring their roots.</p>
Step-by-Step Solution
Key Concept: If a quadratic equation with roots r and s remains unchanged after squaring the roots, then the equation with roots r² and s² must be identical to the original equation. This means {r, s} = {r², s²} as sets, leading to specific relationships between the roots.
<p><strong>Step 1:</strong> Let the quadratic equation be x² + bx + c = 0 with roots r and s. The equation remains unchanged after squaring roots means the equation with roots r² and s² is identical.</p><p><strong>Step 2:</strong> This requires {r, s} = {r², s²} as multisets. We have two cases:</p><p><strong>Case 1:</strong> r = r² and s = s² → r ∈ {0, 1} and s ∈ {0, 1}</p><p>This gives equations: x² = 0, x(x-1) = 0, and x² - x = 0</p><p><strong>Case 2:</strong> r = s² and s = r² (roots swap)</p><p>From r = s² and s = r²: r = (r²)² = r⁴, so r⁴ - r = 0 → r(r³ - 1) = 0</p><p>Thus r ∈ {0, 1, ω, ω²} where ω is a complex cube root of unity.</p><p>For real roots: r = 0 gives s = 0 (already counted) or r = 1 gives s = 1 (already counted)</p><p><strong>Step 3:</strong> Distinct quadratic equations with real roots:</p><p>1. x² = 0 (roots: 0, 0)</p><p>2. x² - x = 0 (roots: 0, 1)</p><p>3. x² - 2x + 1 = 0 or (x-1)² = 0 (roots: 1, 1)</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3