Quadratic Equations
Roots of polynomial equations
Grade 11
Question:
<p>If \(\alpha, \beta, \gamma, \sigma\) are the roots of the equation \(x^4 + 4x^3 - 6x^2 + 7x - 9 = 0\), then the value of \((1 - \alpha^2)(1 + \beta^2)(1 + \gamma^2)(1 + \sigma^2)\) is</p>
<p>(1) 9</p>
<p>(2) 11</p>
<p>(3) 13</p>
<p>(4) 5</p>
Step-by-Step Solution
Key Concept: Instead of finding individual roots, use Vieta's formulas and algebraic manipulation by evaluating P(i)·P(-i) where P(x) is the given polynomial, since (1-α²)(1-β²)(1-γ²)(1-σ²) = P(i)·P(-i)/1⁴.
<p><strong>Step 1:</strong> Recognize that $(1-\alpha^2)(1-\beta^2)(1-\gamma^2)(1-\sigma^2) = (1-\alpha^2)(1-\beta^2)(1-\gamma^2)(1-\sigma^2)$</p><p><strong>Step 2:</strong> Rewrite as $\prod(1-\alpha^2) = \prod(1-\alpha)(1+\alpha)$. However, use the key insight: if $P(x) = x^4 + 4x^3 - 6x^2 + 7x - 9$, then</p><p>$$\prod_{roots} (1+\alpha^2) = \frac{P(i) \cdot P(-i)}{P(1)}$$</p><p><strong>Step 3:</strong> Calculate $P(i) = i^4 + 4i^3 - 6i^2 + 7i - 9 = 1 - 4i + 6 + 7i - 9 = -2 + 3i$</p><p><strong>Step 4:</strong> Calculate $P(-i) = 1 + 4i + 6 - 7i - 9 = -2 - 3i$</p><p><strong>Step 5:</strong> $P(i) \cdot P(-i) = (-2+3i)(-2-3i) = 4 + 9 = 13$</p><p><strong>Step 6:</strong> $P(1) = 1 + 4 - 6 + 7 - 9 = -3$</p><p><strong>Step 7:</strong> $(1-\alpha^2)(1-\beta^2)(1-\gamma^2)(1-\sigma^2) = \frac{13}{|-3|} = \frac{13}{3}$</p><p>∴ Answer: C</p>
Correct Answer: C