Quadratic Equations
Cubic Equations and Roots
Grade 11
Question:
<p>Given that <i>α, β, γ</i> are all real roots of the equation <i>x</i><sup>3</sup> − 2007<i>x</i> + 2002 = 0, then the value of <span>\(\frac{α − 1}{α + 1} + \frac{β − 1}{β + 1} + \frac{γ − 1}{γ + 1}\)</span> is equal to</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to relate the roots to coefficients, then express the sum of fractions in terms of elementary symmetric polynomials of the roots.
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>For the equation x³ - 2007x + 2002 = 0, comparing with x³ + bx² + cx + d = 0:</p><p>• α + β + γ = 0 (coefficient of x² is 0)</p><p>• αβ + βγ + γα = -2007 (coefficient of x)</p><p>• αβγ = -2002 (constant term with sign change)</p><p><strong>Step 2: Simplify the Required Expression</strong></p><p>Let S = (α - 1)/(α + 1) + (β - 1)/(β + 1) + (γ - 1)/(γ + 1)</p><p>Combine over a common denominator:</p><p>S = [(α - 1)(β + 1)(γ + 1) + (β - 1)(α + 1)(γ + 1) + (γ - 1)(α + 1)(β + 1)] / [(α + 1)(β + 1)(γ + 1)]</p><p><strong>Step 3: Calculate the Denominator</strong></p><p>(α + 1)(β + 1)(γ + 1) = αβγ + αβ + βγ + γα + α + β + γ + 1</p><p>= -2002 + (-2007) + 0 + 1</p><p>= -4008</p><p><strong>Step 4: Calculate the Numerator</strong></p><p>Expand each product:</p><p>(α - 1)(β + 1)(γ + 1) = (α - 1)[βγ + β + γ + 1]</p><p>= αβγ + αβ + αγ + α - βγ - β - γ - 1</p><p>Similarly for the other two terms. Adding all three:</p><p>Numerator = αβγ + αβ + αγ + α - βγ - β - γ - 1</p><p>+ αβγ + βα + βγ + β - αγ - α - γ - 1</p><p>+ αβγ + γα + γβ + γ - αβ - α - β - 1</p><p>= 3αβγ + (αβ + βα - αβ) + (αγ - αγ + γα) + (βγ - βγ + γβ) + (α + β + γ - α - α - β - b) + (-3)</p><p>= 3αβγ + αβ + αγ + βγ + (α + β + γ - 2α - 2β - 2γ) - 3</p><p>= 3αβγ + αβ + βγ + γα - (α + β + γ) - 3</p><p>= 3(-2002) + (-2007) - 0 - 3</p><p>= -6006 - 2007 - 3</p><p>= -8016</p><p><strong>Step 5: Calculate the Final Answer</strong></p><p>S = -8016 / (-4008) = 8016 / 4008 = 2</p><p><strong>∴ Answer: 2</strong></p>
Correct Answer: 2