Quadratic Equations
Algebraic identities and equations
Grade 11
Question:
<p>If \(8\alpha^3 + \beta^3 - \gamma^3 + 6\alpha\beta\gamma = 0\) and \(\alpha^2 + 3\gamma = 2\beta\) where \(\alpha,\ \beta,\ \gamma \in R\) and \(\beta + \gamma \neq 0\), then find the largest integral value of \(\gamma\).</p>
Step-by-Step Solution
Key Concept: Recognize that 8α³ + β³ - γ³ + 6αβγ = 0 factors as (2α + β - γ)(4α² + β² + γ² - 2αβ - 2βγ + 2αγ) = 0. Combined with α² + 3γ = 2β, deduce that 2α + β = γ, then substitute to find a constraint on γ.
<p><strong>Step 1:</strong> Factor the equation 8α³ + β³ - γ³ + 6αβγ = 0 using the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).</p><p>Rewrite as (2α)³ + β³ + (-γ)³ - 3(2α)(β)(-γ) = 0, which factors as:</p><p>(2α + β - γ)[(2α)² + β² + γ² - 2αβ + 2αγ - βγ] = 0</p><p><strong>Step 2:</strong> The second factor is always non-negative (sum of squares form). Since β + γ ≠ 0 rules out 2α = β = γ, we must have:</p><p>2α + β - γ = 0 ⟹ γ = 2α + β</p><p><strong>Step 3:</strong> Substitute into α² + 3γ = 2β:</p><p>α² + 3(2α + β) = 2β</p><p>α² + 6α + 3β = 2β</p><p>α² + 6α + β = 0 ⟹ β = -α² - 6α</p><p><strong>Step 4:</strong> Find γ:</p><p>γ = 2α + β = 2α + (-α² - 6α) = -α² - 4α</p><p><strong>Step 5:</strong> To maximize γ, complete the square:</p><p>γ = -(α² + 4α) = -(α² + 4α + 4 - 4) = -(α + 2)² + 4</p><p>Maximum value of γ is 4 when α = -2.</p><p>Since γ must be real and -(α + 2)² + 4 ≤ 4 for all real α, the largest integral value is:</p><p><strong>∴ Answer: 4</strong></p>
Correct Answer: 4