Quadratic Equations
Nature of Roots
Grade 11
Question:
<p>For \(x^2 - (a+3)|x| + 4 = 0\) to have real solutions, the range of \(a\) is</p>
<p>(1) \((-\infty, -7] \cup [1, \infty)\)</p>
<p>(2) \((-3, \infty)\)</p>
<p>(3) \((-\infty, -7]\)</p>
<p>(4) \([1, \infty)\)</p>
Step-by-Step Solution
Key Concept: Since the equation contains |x|, if x = k is a solution, then x = -k is also a solution. Substitute y = |x| ≥ 0 to convert this into a standard quadratic: y² - (a+3)y + 4 = 0, which must have at least one non-negative real root.
<p><strong>Step 1:</strong> Substitute y = |x| where y ≥ 0. The equation becomes: y² - (a+3)y + 4 = 0</p><p><strong>Step 2:</strong> For the original equation to have real solutions, this quadratic in y must have at least one non-negative real root.</p><p><strong>Step 3:</strong> Let f(y) = y² - (a+3)y + 4. For at least one non-negative root:</p><p>• Product of roots = 4 > 0 (always positive)</p><p>• Sum of roots = a + 3</p><p><strong>Step 4:</strong> Since the product is positive, both roots have the same sign. For at least one non-negative root, both must be non-negative, requiring:</p><p>• Sum of roots ≥ 0: a + 3 ≥ 0 ⟹ a ≥ -3</p><p>• Discriminant ≥ 0: (a+3)² - 16 ≥ 0 ⟹ (a+3)² ≥ 16</p><p><strong>Step 5:</strong> From (a+3)² ≥ 16: |a+3| ≥ 4</p><p>⟹ a + 3 ≥ 4 or a + 3 ≤ -4</p><p>⟹ a ≥ 1 or a ≤ -7</p><p><strong>Step 6:</strong> Combining with a ≥ -3: we need a ≥ 1 or (a ≤ -7 ∩ a ≥ -3) = ∅</p><p>∴ Answer: <strong>a ∈ [1, ∞)</strong> or <strong>a ≥ 1</strong></p>
Correct Answer: A