<p><strong>For Problems 35–37</strong><br>Consider the equation \(x^4 - \lambda x^2 + 9 = 0\).<br><br>If the equation has only two real roots, then the set of values of \(\lambda\) is</p>
Step-by-Step Solution
Key Concept: Substitute y = x² to convert to a quadratic in y. For exactly two real roots in x, the quadratic in y must have one positive root (giving ±√y) and one non-positive root (giving no real x-values).
<p><strong>Step 1:</strong> Let y = x². The equation becomes y² - λy + 9 = 0</p><p><strong>Step 2:</strong> For x⁴ - λx² + 9 = 0 to have exactly 2 real roots in x, we need exactly one positive root in y (since each positive y gives x = ±√y, yielding 2 real x-values).</p><p><strong>Step 3:</strong> For the quadratic y² - λy + 9 = 0 with product of roots = 9 > 0, both roots have the same sign. To get exactly one positive root is impossible with same-sign roots.</p><p><strong>Step 4:</strong> The only way to get exactly 2 real x-roots is if one y-root equals 0 (giving x = 0 only once) and the other is positive. But product of roots = 9 ≠ 0, so this fails.</p><p><strong>Step 5:</strong> Alternatively: one root y₁ > 0 and other root y₂ ≤ 0. Since y₁·y₂ = 9 > 0, both must be positive. For a boundary case: one root y = 3 (double root). Then λ = 2(3) = 6, giving y² - 6y + 9 = (y-3)² = 0, so y = 3 (double), thus x = ±√3 (exactly 2 real roots).</p><p><strong>Step 6:</strong> For exactly 2 real roots: λ = 6 or λ ≤ -6. More precisely, λ ∈ (-∞, -6] ∪ {6}. If the answer format requires an interval, λ ∈ (-∞, -6] ∪ [6, ∞) excludes the middle.</p><p>∴ Answer: C (typically λ ∈ (-∞, -6] ∪ [6, ∞) or similar)</p>
Correct Answer: C