Quadratic Equations
Formation of quadratic equations
Grade 11
Question:
<p>Let <i>p</i> and <i>q</i> be real numbers such that <i>p</i> ≠ 0, <i>p</i><sup>3</sup> ≠ <i>q</i>, and <i>p</i><sup>3</sup> ≠ −<i>q</i>. If α and β are nonzero complex numbers satisfying α + β = −<i>p</i> and α<sup>3</sup> + β<sup>3</sup> = <i>q</i>, then a quadratic equation having α/β and β/α as its roots is</p>
<p>\((p^3 + q)x^2 - (p^3 + 2q)x + (p^3 + q) = 0\)</p>
<p>\((p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0\)</p>
<p>\((p^3 - q)x^2 - (5p^3 - 2q)x + (p^3 - q) = 0\)</p>
<p>\((p^3 - q)x^2 - (5p^3 + 2q)x + (p^3 - q) = 0\)</p>
Step-by-Step Solution
Key Concept: Use the identity α³ + β³ = (α + β)³ - 3αβ(α + β) to find αβ, then construct a quadratic with roots α/β and β/α using sum = (α² + β²)/(αβ) and product = 1.
<p><strong>Step 1: Find αβ using the given identity</strong></p><p>We know: α + β = -p and α³ + β³ = q</p><p>Using α³ + β³ = (α + β)³ - 3αβ(α + β):</p><p>q = (-p)³ - 3αβ(-p)</p><p>q = -p³ + 3pαβ</p><p>Therefore: αβ = (q + p³)/(3p)</p><p><strong>Step 2: Find sum of roots α/β + β/α</strong></p><p>α/β + β/α = (α² + β²)/(αβ)</p><p>Since α² + β² = (α + β)² - 2αβ = p² - 2(q + p³)/(3p)</p><p>α² + β² = p² - (2q + 2p³)/(3p) = (3p³ - 2q - 2p³)/(3p) = (p³ - 2q)/(3p)</p><p>Sum = [(p³ - 2q)/(3p)] ÷ [(q + p³)/(3p)] = (p³ - 2q)/(q + p³)</p><p><strong>Step 3: Find product of roots</strong></p><p>Product: (α/β) · (β/α) = 1</p><p><strong>Step 4: Form the quadratic equation</strong></p><p>With roots having sum S = (p³ - 2q)/(q + p³) and product P = 1:</p><p>x² - Sx + P = 0</p><p>x² - [(p³ - 2q)/(q + p³)]x + 1 = 0</p><p>Multiply by (q + p³): (q + p³)x² - (p³ - 2q)x + (q + p³) = 0</p><p>∴ Answer: B</p>
Correct Answer: B