Quadratic Equations
Roots of quadratic equations and substitution
Grade 11
Question:
<p>If \(\alpha\) is the root of the equation \(x^2 - x + 2 = 0\), then the value of \(\dfrac{6(-\alpha^3 + 2\alpha^2 - \alpha)}{\alpha^5 - 3\alpha^4 + 3\alpha^3 - \alpha^2}\) is equal to:</p>
<p>3</p>
<p>6</p>
<p>9</p>
<p>12</p>
Step-by-Step Solution
Key Concept: Since α is a root of x² - x + 2 = 0, we have α² = α - 2. Use this relation to reduce all higher powers of α to linear expressions, then simplify the given fraction.
<p><strong>Step 1:</strong> Since α is a root of x² - x + 2 = 0, we have <strong>α² = α - 2</strong></p><p><strong>Step 2:</strong> Reduce higher powers using α² = α - 2:</p><p>α³ = α·α² = α(α - 2) = α² - 2α = (α - 2) - 2α = -α - 2</p><p>α⁴ = α·α³ = α(-α - 2) = -α² - 2α = -(α - 2) - 2α = -3α + 2</p><p>α⁵ = α·α⁴ = α(-3α + 2) = -3α² + 2α = -3(α - 2) + 2α = -α + 6</p><p><strong>Step 3:</strong> Simplify the numerator:</p><p>-α³ + 2α² - α = -(-α - 2) + 2(α - 2) - α = α + 2 + 2α - 4 - α = 2α - 2 = 2(α - 1)</p><p>Numerator: 6·2(α - 1) = 12(α - 1)</p><p><strong>Step 4:</strong> Simplify the denominator:</p><p>α⁵ - 3α⁴ + 3α³ - α² = (-α + 6) - 3(-3α + 2) + 3(-α - 2) - (α - 2)</p><p>= -α + 6 + 9α - 6 - 3α - 6 - α + 2 = 4α - 4 = 4(α - 1)</p><p><strong>Step 5:</strong> Divide:</p><p>$$\frac{12(α - 1)}{4(α - 1)} = \frac{12}{4} = 3$$</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: B