Quadratic Equations
Equations involving modulus
Grade 11
Question:
<p>84. 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|, substitute y = |x| ≥ 0 to get y² - (a+3)y + 4 = 0. For real solutions in x, this quadratic in y 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 real solutions in x, we need at least one non-negative root y ≥ 0.</p><p><strong>Step 3:</strong> Let f(y) = y² - (a+3)y + 4. Product of roots = 4 > 0, so both roots have the same sign.</p><p><strong>Step 4:</strong> Since product > 0 and we need at least one non-negative root, both roots must be positive. This requires:</p><p>• Discriminant: (a+3)² - 16 ≥ 0 → (a+3)² ≥ 16 → a+3 ≥ 4 or a+3 ≤ -4 → a ≥ 1 or a ≤ -7</p><p>• Sum of roots > 0: a+3 > 0 → a > -3</p><p><strong>Step 5:</strong> Combining conditions: (a ≥ 1 or a ≤ -7) AND a > -3</p><p>Only a ≥ 1 satisfies both conditions.</p><p>∴ Answer: a ∈ [1, ∞)</p>
Correct Answer: 1