Quadratic Equations
Symmetric functions of roots
Grade 11
Question:
<p>If \(\alpha, \beta\) are the roots of \(x^2 - px + q = 0\) and \(\alpha', \beta'\) are the roots of \(x^2 - p'x + q' = 0\), then the value of \((\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2\) is</p>
<p>\(2\{p^2 - 2q + p'^2 - 2q' - pp'\}\)</p>
<p>\(2\{p^2 - 2q + p'^2 - 2q' + qq'\}\)</p>
<p>\(2\{p^2 - 2q - p'^2 - 2q' + pp'\}\)</p>
<p>\(2\{p^2 - 2q - p'^2 - 2q' - qq'\}\)</p>
Step-by-Step Solution
Key Concept: Expand the expression using Vieta's formulas (α + β = p, αβ = q, α' + β' = p', α'β' = q') and recognize that the sum simplifies to 2[(α + β)² + (α' + β')² - 2(αα' + αβ' + βα' + ββ')] = 2[(α-α')² + (α-β')² + (β-α')² + (β-β')²] can be rewritten using sum of squares identities.
<p><strong>Step 1:</strong> Let S = (α - α')² + (β - α')² + (α - β')² + (β - β')²</p><p><strong>Step 2:</strong> Expand each term:<br/>= α² - 2αα' + α'² + β² - 2βα' + α'² + α² - 2αβ' + β'² + β² - 2ββ' + β'²<br/>= 2(α² + β²) + 2(α'² + β'²) - 2α(α' + β') - 2β(α' + β')</p><p><strong>Step 3:</strong> Factor out and use Vieta's formulas:<br/>= 2(α² + β²) + 2(α'² + β'²) - 2(α + β)(α' + β')<br/>= 2[(α + β)² - 2αβ] + 2[(α' + β')² - 2α'β'] - 2(α + β)(α' + β')<br/>= 2[p² - 2q] + 2[p'² - 2q'] - 2pp'</p><p><strong>Step 4:</strong> Simplify:<br/>= 2p² - 4q + 2p'² - 4q' - 2pp'<br/>= 2(p² + p'² - pp' - 2q - 2q')</p><p>∴ Answer: <strong>2(p² + p'² - pp' - 2q - 2q')</strong> or equivalently <strong>2[(p - p')² + pp' - 2(q + q')]</strong> (Option A)</p>
Correct Answer: A