Quadratic Equations
Roots and coefficients
Grade 11
Question:
<p>If \(\alpha\) and \(\beta\) are roots of \(x^2 - 2px + p^2 - 1 = 0\) and \(\left|\frac{\alpha^2+\beta^2}{\alpha\beta}+3\right| \leq 5\), then \(p \in [a,\,b]\). Find the value of \([2(a^2+b^2)]\).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express α²+β² and αβ in terms of p, then simplify the absolute value inequality to find the range of p. The expression (α²+β²)/αβ = (α+β)²/αβ - 2 reduces to a polynomial inequality in p.
<p><strong>Step 1:</strong> Apply Vieta's formulas to x² - 2px + p² - 1 = 0</p><p>α + β = 2p and αβ = p² - 1</p><p><strong>Step 2:</strong> Simplify the given expression</p><p>α² + β² = (α + β)² - 2αβ = 4p² - 2(p² - 1) = 2p² + 2</p><p>Therefore: (α² + β²)/(αβ) = (2p² + 2)/(p² - 1) = 2(p² + 1)/(p² - 1)</p><p><strong>Step 3:</strong> Rewrite the absolute value inequality</p><p>|2(p² + 1)/(p² - 1) + 3| ≤ 5</p><p>|[2(p² + 1) + 3(p² - 1)]/(p² - 1)| ≤ 5</p><p>|(5p² - 1)/(p² - 1)| ≤ 5</p><p><strong>Step 4:</strong> Solve the compound inequality -5 ≤ (5p² - 1)/(p² - 1) ≤ 5</p><p><u>Left inequality:</u> (5p² - 1)/(p² - 1) ≥ -5</p><p>(5p² - 1 + 5p² - 5)/(p² - 1) ≥ 0 → (10p² - 6)/(p² - 1) ≥ 0</p><p><u>Right inequality:</u> (5p² - 1)/(p² - 1) ≤ 5</p><p>(5p² - 1 - 5p² + 5)/(p² - 1) ≤ 0 → 4/(p² - 1) ≤ 0</p><p>This requires p² < 1, so -1 < p < 1</p><p><strong>Step 5:</strong> Combine both inequalities and note p² ≠ 1</p><p>From right inequality: -1 < p < 1</p><p>Check left inequality for this range: both conditions give p ∈ [-1, 1]</p><p>Therefore a = -1, b = 1</p><p><strong>Step 6:</strong> Calculate final answer</p><p>[2(a² + b²)] = [2(1 + 1)] = [4] = <strong>4</strong></p>
Correct Answer: 2