<p>Let \(\alpha\) and \(\beta\) are two roots of the equation \(x^2 + px + q = 0\), where \(p\) and \(q\) are real numbers, and \(q \neq 0\). Now suppose another quadratic equation \(x^2 + mx + n = 0\) with roots \(\alpha + \dfrac{1}{\alpha}\) and \(\beta + \dfrac{1}{\beta}\) such that \(m + n = 0\). Then the possible integral values in the range of \(q\) can be:</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to relate the roots of the original equation to the new equation, then apply the constraint m + n = 0 to establish bounds on q. The key is finding the sum and product of the new roots in terms of p and q.
<p><strong>Step 1: Set up using Vieta's formulas for the original equation.</strong></p><p>From x² + px + q = 0 with roots α and β:</p><p>α + β = -p and αβ = q</p><p><strong>Step 2: Find the sum of roots of the new equation.</strong></p><p>Sum of new roots = (α + 1/α) + (β + 1/β)</p><p>= α + β + (1/α + 1/β)</p><p>= α + β + (α + β)/(αβ)</p><p>= -p + (-p)/q = -p(1 + 1/q) = -p(q+1)/q</p><p>By Vieta's: Sum = -m, so m = p(q+1)/q</p><p><strong>Step 3: Find the product of roots of the new equation.</strong></p><p>Product = (α + 1/α)(β + 1/β)</p><p>= αβ + α/β + β/α + 1/(αβ)</p><p>= αβ + (α² + β²)/(αβ) + 1/(αβ)</p><p>= q + ((α+β)² - 2αβ)/(αβ) + 1/q</p><p>= q + (p² - 2q)/q + 1/q</p><p>= q + (p² - 2q + 1)/q = (q² + p² - 2q + 1)/q</p><p>By Vieta's: Product = n, so n = (q² + p² - 2q + 1)/q</p><p><strong>Step 4: Apply the condition m + n = 0.</strong></p><p>m + n = 0</p><p>p(q+1)/q + (q² + p² - 2q + 1)/q = 0</p><p>p(q+1) + q² + p² - 2q + 1 = 0</p><p>pq + p + q² + p² - 2q + 1 = 0</p><p>p² + pq + p + q² - 2q + 1 = 0</p><p><strong>Step 5: Treat as a quadratic in p and find the discriminant condition.</strong></p><p>p² + p(q+1) + (q² - 2q + 1) = 0</p><p>p² + p(q+1) + (q-1)² = 0</p><p>For real p, Δ ≥ 0:</p><p>(q+1)² - 4(q-1)² ≥ 0</p><p>(q+1)² ≥ 4(q-1)²</p><p>|q+1| ≥ 2|q-1|</p><p><strong>Step 6: Solve the inequality.</strong></p><p>Case 1: q+1 ≥ 2(q-1) gives q+1 ≥ 2q-2, so 3 ≥ q, hence q ≤ 3</p><p>Case 2: q+1 ≥ -2(q-1) gives q+1 ≥ -2q+2, so 3q ≥ 1, hence q ≥ 1/3</p><p>Combined with q ≠ 0: 1/3 ≤ q ≤ 3</p><p><strong>Step 7: Find integral values in this range.</strong></p><p>The range is [1/3, 3]. The integral values are: 1 and 2 and 3</p><p>However, checking q = 3: (3+1)² = 16 and 4(3-1)² = 16, so equality holds (p = 0, valid)</p><p>∴ Answer: A,B</p>
Correct Answer: A,B