<p>Find the sum of all integral values of \(a\) for which all the roots of the equation \(x^4 - 4x^3 - 8x^2 + a = 0\) are real.</p>
Step-by-Step Solution
Key Concept: For all roots of a quartic to be real, we need to analyze when the equation can be factored into two quadratics with real roots. Rewrite as x⁴ - 4x³ - 8x² + a = 0 and use the condition that the discriminant structure allows real roots.
<p><strong>Step 1:</strong> Rewrite the equation as x⁴ - 4x³ - 8x² + a = 0. Complete the square by writing it in terms of (x²).</p><p><strong>Step 2:</strong> Let f(x) = x⁴ - 4x³ - 8x². We need f(x) = -a to have all real roots. Note that f(x) = x²(x² - 4x - 8).</p><p><strong>Step 3:</strong> Find critical points of f(x): f'(x) = 4x³ - 12x² - 16x = 4x(x² - 3x - 4) = 4x(x-4)(x+1). Critical points at x = -1, 0, 4.</p><p><strong>Step 4:</strong> Evaluate f at critical points:</p><p>f(-1) = 1 + 4 - 8 = -3</p><p>f(0) = 0</p><p>f(4) = 256 - 256 - 128 = -128</p><p><strong>Step 5:</strong> The minimum value of f(x) is -128 (at x = 4). For all roots to be real, we need -a ≥ -128, so a ≤ 128.</p><p><strong>Step 6:</strong> The maximum occurs at the local maximum. As x → ∞, f(x) → ∞. The function has local maximum at x = -1 where f(-1) = -3. For real roots to exist, we need -a ≤ 0 (approximately), giving a ≥ 0.</p><p><strong>Step 7:</strong> More precisely, all roots are real when -128 ≤ -a ≤ 0, which means 0 ≤ a ≤ 128.</p><p><strong>Step 8:</strong> Integral values: a ∈ {0, 1, 2, ..., 128}. Sum = 0 + 1 + 2 + ... + 128 = (128 × 129)/2 = 8256.</p><p>∴ Answer: <strong>8256</strong></p>
Correct Answer: 0