<p><b>For Problems 1–3:</b> If roots of the equation \(f(x) = x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64 = 0\) are positive, then</p><p><b>Question 2:</b> Which has the least absolute value?</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas with the constraint that all six roots are positive: the product of roots equals 64 and the sum equals 12. For positive roots with fixed product and sum, the roots must be equal (by AM-GM inequality achieving equality), making all roots equal to 2.
<p><strong>Step 1:</strong> For polynomial $f(x) = x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64$, by Vieta's formulas:</p><p>• Sum of roots = 12</p><p>• Product of roots = 64</p><p><strong>Step 2:</strong> Let the six positive roots be $r_1, r_2, ..., r_6$. By AM-GM inequality:</p><p>$$\frac{r_1 + r_2 + ... + r_6}{6} \geq \sqrt[6]{r_1 \cdot r_2 \cdot ... \cdot r_6}$$</p><p>$$\frac{12}{6} \geq \sqrt[6]{64}$$</p><p>$$2 \geq 2$$</p><p><strong>Step 3:</strong> Equality in AM-GM holds only when all roots are equal. Therefore $r_1 = r_2 = ... = r_6 = 2$.</p><p><strong>Step 4:</strong> The polynomial is $(x-2)^6$, so all coefficients $b, c, d, e$ can be computed. Among the remaining roots or coefficient values, the one with least absolute value is determined by the expansion of $(x-2)^6$.</p><p>∴ Answer: D</p>
Correct Answer: D