<p>If \((x-2)^6 + (x-4)^6 = 64\), then equation has</p>
<p>(a) real and irrational roots</p>
<p>(b) real and rational roots</p>
<p>(c) two real and two irrational roots</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Substitute y = x - 3 to center the equation at the midpoint between 2 and 4, transforming it into (y+1)^6 + (y-1)^6 = 64, which eliminates odd powers and reveals the structure. This symmetric substitution dramatically simplifies the problem.
<p><strong>Step 1:</strong> Let y = x - 3, so x - 2 = y + 1 and x - 4 = y - 1</p><p><strong>Step 2:</strong> Substitute: (y+1)⁶ + (y-1)⁶ = 64</p><p><strong>Step 3:</strong> Expand using binomial theorem. Since we're adding, odd powers cancel:</p><p>(y+1)⁶ + (y-1)⁶ = 2[y⁶ + 15y⁴ + 15y² + 1]</p><p><strong>Step 4:</strong> So: 2[y⁶ + 15y⁴ + 15y² + 1] = 64</p><p>y⁶ + 15y⁴ + 15y² + 1 = 32</p><p>y⁶ + 15y⁴ + 15y² - 31 = 0</p><p><strong>Step 5:</strong> Let z = y². Then: z³ + 15z² + 15z - 31 = 0</p><p><strong>Step 6:</strong> Testing z = 1: 1 + 15 + 15 - 31 = 0 ✓</p><p><strong>Step 7:</strong> Factor: (z - 1)(z² + 16z + 31) = 0</p><p>z = 1 or z² + 16z + 31 = 0</p><p><strong>Step 8:</strong> For z² + 16z + 31 = 0: Discriminant = 256 - 124 = 132 > 0 gives z = -8 ± √33 (both negative, so no real y)</p><p><strong>Step 9:</strong> Only z = 1 (i.e., y² = 1) gives real solutions: y = ±1</p><p>Therefore: x = 3 ± 1, so x = 4 or x = 2</p><p>∴ The equation has <strong>2 real roots</strong></p>
Correct Answer: B