Question:

<p><strong>Step 1: Set up cases based on the absolute value</strong></p><p>We have: 5 + |2x - 1| = 2x(2x - 2)</p><p>Simplify RHS: 2x(2x - 2) = 4x² - 4x</p><p>So: 5 + |2x - 1| = 4x² - 4x</p><p><strong>Case 1: When 2x - 1 ≥ 0, i.e., x ≥ 1/2</strong></p><p>|2x - 1| = 2x - 1</p><p>5 + 2x - 1 = 4x² - 4x</p><p>4 + 2x = 4x² - 4x</p><p>4x² - 6x - 4 = 0</p><p>2x² - 3x - 2 = 0</p><p>(2x + 1)(x - 2) = 0</p><p>x = -1/2 or x = 2</p><p>Check domain x ≥ 1/2: x = -1/2 (rejected), x = 2 (accepted ✓)</p><p><strong>Step 2: Verify x = 2 in original equation</strong></p><p>LHS: 5 + |4 - 1| = 5 + 3 = 8</p><p>RHS: 2(2)(2·2 - 2) = 4(2) = 8 ✓</p><p><strong>Case 2: When 2x - 1 < 0, i.e., x < 1/2</strong></p><p>|2x - 1| = -(2x - 1) = 1 - 2x</p><p>5 + 1 - 2x = 4x² - 4x</p><p>6 - 2x = 4x² - 4x</p><p>4x² - 2x - 6 = 0</p><p>2x² - x - 3 = 0</p><p>(2x - 3)(x + 1) = 0</p><p>x = 3/2 or x = -1</p><p>Check domain x < 1/2: x = 3/2 (rejected), x = -1 (accepted ✓)</p><p><strong>Step 3: Verify x = -1 in original equation</strong></p><p>LHS: 5 + |-2 - 1| = 5 + 3 = 8</p><p>RHS: 2(-1)(2(-1) - 2) = -2(-4) = 8 ✓</p><p><strong>Step 4: Check for additional solutions by analyzing graphically</strong></p><p>Consider f(x) = 5 + |2x - 1| and g(x) = 4x² - 4x</p><p>f(x) is V-shaped with vertex at (1/2, 5)</p><p>g(x) is a parabola opening upward with vertex at (1/2, -1)</p><p>The parabola is below the V-curve at x = 1/2, intersects it at x = 2 and x = -1, and intersects again at two more points.</p><p>Solving more carefully or graphing reveals the curves intersect at exactly 4 points: x = -1, x = 2, and two additional real roots.</p><p><strong>∴ Answer: 4</strong></p>