Quadratic Equations
Transformation of roots
Grade 11
Question:
<p>If \(\alpha, \beta, \gamma\) are the roots of \(x^3 - x^2 - 1 = 0\), then the value of \(\dfrac{1+\alpha}{1-\alpha} + \dfrac{1+\beta}{1-\beta} + \dfrac{1+\gamma}{1-\gamma}\) is equal to</p>
<p>(1) -5</p>
<p>(2) -6</p>
<p>(3) -7</p>
<p>(4) -2</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find the symmetric sum of roots, then convert the expression into a rational form that depends only on elementary symmetric polynomials of α, β, γ.
<p><strong>Step 1:</strong> From Vieta's formulas for $x^3 - x^2 - 1 = 0$:</p><p>$\alpha + \beta + \gamma = 1$</p><p>$\alpha\beta + \beta\gamma + \gamma\alpha = 0$</p><p>$\alpha\beta\gamma = 1$</p><p><strong>Step 2:</strong> Let $S = \frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$</p><p>Find common denominator: $S = \frac{(1+\alpha)(1-\beta)(1-\gamma) + (1+\beta)(1-\alpha)(1-\gamma) + (1+\gamma)(1-\alpha)(1-\beta)}{(1-\alpha)(1-\beta)(1-\gamma)}$</p><p><strong>Step 3:</strong> Calculate denominator:</p><p>$(1-\alpha)(1-\beta)(1-\gamma) = 1 - (\alpha+\beta+\gamma) + (\alpha\beta+\beta\gamma+\gamma\alpha) - \alpha\beta\gamma$</p><p>$= 1 - 1 + 0 - 1 = -1$</p><p><strong>Step 4:</strong> Expand numerator using the identity pattern:</p><p>Numerator $= 3 + 2(\alpha+\beta+\gamma) - (\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$</p><p>$= 3 + 2(1) - 0 - 3(1) = 3 + 2 - 3 = 2$</p><p><strong>Step 5:</strong> Therefore: $S = \frac{2}{-1} = -2$</p><p>∴ Answer: C</p>
Correct Answer: C