Basic Mathematics & Logarithm
Absolute value equations and inequalities
Grade 11
Question:
<p>The value of \(x\) for which the equation \(|x^2 + 6x + 6| = |x^2 + 4x + 9| + |2x - 3|\) holds, is equal to:</p>
<p>(a) \(\left[\dfrac{3}{2}, \infty\right)\)</p>
<p>(b) \(\left(-\infty, \dfrac{3}{2}\right]\)</p>
<p>(c) \((-\infty, 1] \cup \left[\dfrac{3}{2}, \infty\right)\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: For an equation involving absolute values to hold, we need |A| = |B| + |C|. This is only possible when B and C have specific sign relationships—particularly when |B| + |C| equals |A| at critical points where expressions change sign. Test the boundary points where each expression inside absolute values equals zero.
<p><strong>Step 1:</strong> Identify critical points where expressions equal zero.</p><p>For x² + 6x + 6 = 0: x = -3 ± √3</p><p>For x² + 4x + 9 = 0: Δ = 16 - 36 = -20 < 0 (always positive)</p><p>For 2x - 3 = 0: x = 3/2</p><p><strong>Step 2:</strong> Use the constraint that |A| = |B| + |C|. Since x² + 4x + 9 > 0 always, we have |x² + 4x + 9| = x² + 4x + 9.</p><p><strong>Step 3:</strong> Test x = 3/2 (critical point of linear term).</p><p>LHS: |(3/2)² + 6(3/2) + 6| = |9/4 + 9 + 6| = |9/4 + 15| = 69/4</p><p>RHS: |(3/2)² + 4(3/2) + 9| + |2(3/2) - 3| = |9/4 + 6 + 9| + |3 - 3| = 69/4 + 0 = 69/4</p><p><strong>Step 4:</strong> Verify LHS = RHS at x = 3/2: 69/4 = 69/4 ✓</p><p>∴ Answer: x = 3/2 or B</p>
Correct Answer: B