<p><strong>164.</strong> If the roots of \(x^4 + qx^2 + kx + 225 = 0\) are in arithmetic progression, then the value of \(q\) is:</p>
Step-by-Step Solution
Key Concept: If roots are in AP, use symmetry properties. For a quartic with roots in AP, the sum of outer roots equals sum of inner roots, allowing you to establish that the coefficient of x³ being zero forces a special relationship between roots.
<p><strong>Step 1:</strong> Let the four roots in AP be (a-3d), (a-d), (a+d), (a+3d), where a is the center and d is common difference.</p><p><strong>Step 2:</strong> Sum of roots = (a-3d)+(a-d)+(a+d)+(a+3d) = 4a. Since coefficient of x³ is 0, we have 4a = 0, so <strong>a = 0</strong>. Roots become: -3d, -d, d, 3d.</p><p><strong>Step 3:</strong> Product of roots = (-3d)(-d)(d)(3d) = 9d⁴ = 225. Thus d⁴ = 25, giving d² = 5 (taking positive value).</p><p><strong>Step 4:</strong> Sum of products of roots taken two at a time: (-3d)(-d) + (-3d)(d) + (-3d)(3d) + (-d)(d) + (-d)(3d) + (d)(3d) = 3d² - 3d² - 9d² - d² - 3d² + 3d² = -10d² = -10(5) = -50.</p><p><strong>Step 5:</strong> By Vieta's formulas, this sum equals q. Therefore <strong>q = -50</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D