<p>If <i>p</i> and <i>q</i> are chosen randomly from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) with replacement, then the probability that the roots of the equation \(x^2 + px + q = 0\)</p>
<p>(1) are real is 33/50</p>
<p>(2) are imaginary is 19/50</p>
<p>(3) are real and equal is 3/50</p>
<p>(4) are real and distinct is 3/5</p>
Step-by-Step Solution
Key Concept: For the quadratic x² + px + q = 0 to have real roots, the discriminant must be non-negative: p² - 4q ≥ 0, which means q ≤ p²/4. Count favorable pairs (p,q) from the 100 total possibilities with replacement.
<p><strong>Step 1:</strong> For real roots of x² + px + q = 0, we need discriminant Δ ≥ 0</p><p>This gives us: p² - 4q ≥ 0, or <strong>q ≤ p²/4</strong></p><p><strong>Step 2:</strong> Count valid pairs (p,q) where p,q ∈ {1,2,3,...,10} with replacement:</p><p>• p=1: q ≤ 0.25 → no valid q</p><p>• p=2: q ≤ 1 → q ∈ {1} → 1 pair</p><p>• p=3: q ≤ 2.25 → q ∈ {1,2} → 2 pairs</p><p>• p=4: q ≤ 4 → q ∈ {1,2,3,4} → 4 pairs</p><p>• p=5: q ≤ 6.25 → q ∈ {1,2,3,4,5,6} → 6 pairs</p><p>• p=6: q ≤ 9 → q ∈ {1,2,...,9} → 9 pairs</p><p>• p=7: q ≤ 12.25 → q ∈ {1,2,...,10} → 10 pairs</p><p>• p=8: q ≤ 16 → q ∈ {1,2,...,10} → 10 pairs</p><p>• p=9: q ≤ 20.25 → q ∈ {1,2,...,10} → 10 pairs</p><p>• p=10: q ≤ 25 → q ∈ {1,2,...,10} → 10 pairs</p><p><strong>Step 3:</strong> Total favorable outcomes = 0+1+2+4+6+9+10+10+10+10 = <strong>62</strong></p><p><strong>Step 4:</strong> Total possible outcomes = 10 × 10 = 100</p><p><strong>Step 5:</strong> Probability = 62/100 = 31/50</p><p>∴ Answer: Probability = 31/50 (or 0.62)</p>
Correct Answer: 2,3,4