Basic Mathematics & Logarithm
Absolute Value Equations/Inequalities
Grade 11
Question:
<p><strong>157.</strong> 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 absolute value equations of the form |A| = |B| + |C|, equality holds only when A, B, C maintain specific sign relationships—specifically when A and B have the same sign and C = 0, or when the expressions are collinear on the number line. Test critical points where expressions change sign.
<p><strong>Step 1: Identify critical points</strong> where expressions change sign:</p><p>• x² + 6x + 6 = 0 at x = -3 ± √3</p><p>• x² + 4x + 9: Discriminant = 16 - 36 = -20 < 0, always positive</p><p>• 2x - 3 = 0 at x = 3/2</p><p><strong>Step 2: Since x² + 4x + 9 > 0 always,</strong> the equation becomes:</p><p>|x² + 6x + 6| = (x² + 4x + 9) + |2x - 3|</p><p><strong>Step 3: Test x = 3/2</strong> (where 2x - 3 = 0):</p><p>LHS: |(9/4) + 9 + 6| = |9/4 + 15| = |69/4| = 69/4</p><p>RHS: (9/4 + 6 + 9) + 0 = 9/4 + 15 = 69/4 ✓</p><p><strong>Step 4: Verify this is the unique solution</strong> by checking that other regions don't satisfy the equation (the coefficient balance breaks down elsewhere).</p><p>∴ Answer: x = 3/2 (Option B)</p>
Correct Answer: B