<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: assume roots as (a-3d, a-d, a+d, a+3d). The sum of roots equals zero (coefficient of x³ is 0), giving a=0, so roots are (-3d, -d, d, 3d). Use Vieta's formulas for the product and x² coefficient.
<p><strong>Step 1:</strong> Since coefficient of x³ is 0, sum of roots = 0. For roots in AP, assume them as (a-3d, a-d, a+d, a+3d).</p><p><strong>Step 2:</strong> Sum: (a-3d)+(a-d)+(a+d)+(a+3d) = 4a = 0 ⟹ a = 0</p><p><strong>Step 3:</strong> Roots are (-3d, -d, d, 3d). Product of roots = (-3d)(-d)(d)(3d) = 9d⁴ = 225 ⟹ d⁴ = 25 ⟹ d² = 5</p><p><strong>Step 4:</strong> Sum of products of roots taken two at a time:</p><p>(-3d)(-d) + (-3d)(d) + (-3d)(3d) + (-d)(d) + (-d)(3d) + (d)(3d)</p><p>= 3d² - 3d² - 9d² - d² - 3d² + 3d²</p><p>= -10d² = -10(5) = -50</p><p>∴ Answer: D (q = -50)</p>
Correct Answer: D