<p>Let three quadratic equations \(ax^2 - 2bx + c = 0\), \(bx^2 - 2cx + a = 0\) and \(cx^2 - 2ax + b = 0\), all have only positive roots. Then which of these are always true?</p>
<p>\(b^2 = ac\)</p>
<p>\(c^2 = ab\)</p>
<p>each pair of equations has exactly one root common</p>
<p>each pair of equations has two roots common</p>
Step-by-Step Solution
Key Concept: For a quadratic equation with only positive roots, both the sum and product of roots must be positive. By applying Vieta's formulas to all three equations and using the constraint that all roots are positive, we can establish relationships between the coefficients a, b, and c.
<p><strong>Step 1: Apply Vieta's formulas to the three equations.</strong></p><p>For equation ax² - 2bx + c = 0: sum of roots = 2b/a, product of roots = c/a</p><p>For equation bx² - 2cx + a = 0: sum of roots = 2c/b, product of roots = a/b</p><p>For equation cx² - 2ax + b = 0: sum of roots = 2a/c, product of roots = b/c</p><p><strong>Step 2: Apply the constraint that all roots are positive.</strong></p><p>Since all roots are positive:</p><p>• From eq(1): 2b/a > 0 and c/a > 0</p><p>• From eq(2): 2c/b > 0 and a/b > 0</p><p>• From eq(3): 2a/c > 0 and b/c > 0</p><p>This implies a, b, c all have the same sign. Assume all positive.</p><p><strong>Step 3: Establish relationship between a, b, c using discriminant condition.</strong></p><p>For real roots, discriminants ≥ 0:</p><p>• Δ₁ = 4b² - 4ac ≥ 0 ⟹ b² ≥ ac</p><p>• Δ₂ = 4c² - 4ab ≥ 0 ⟹ c² ≥ ab</p><p>• Δ₃ = 4a² - 4bc ≥ 0 ⟹ a² ≥ bc</p><p><strong>Step 4: Multiply the three inequalities.</strong></p><p>(b²)(c²)(a²) ≥ (ac)(ab)(bc)</p><p>a²b²c² ≥ a²b²c²</p><p>This is equality, so: b² = ac, c² = ab, and a² = bc (all equalities hold)</p><p><strong>Step 5: Verify options A and B.</strong></p><p>From b² = ac and c² = ab, we get: b⁴ = a²c² = a(ab)c = a²bc = a² · b (since bc = a²)</p><p>This confirms b² = ac and c² = ab are always true. ✓</p><p><strong>Step 6: Test whether equations share common roots.</strong></p><p>Assume equations share a common root α > 0.</p><p>aα² - 2bα + c = 0 ... (1)</p><p>bα² - 2cα + a = 0 ... (2)</p><p>From b² = ac and c² = ab, we can write c = b²/a and verify: if we add appropriately scaled equations, the algebra shows that both roots of one equation satisfy another.</p><p><strong>Step 7: Check for two common roots.</strong></p><p>Since b² = ac and c² = ab imply a³ = b³ = c³ (from cyclic relationships), we have a = b = c.</p><p>When a = b = c, all three equations become x² - 2x + 1 = 0, i.e., (x-1)² = 0.</p><p>Each equation has a double root at x = 1, so each pair shares both roots. ✓</p><p>Option C is false (not exactly one common root in general case).</p><p><strong>∴ Answer: A,B,D</strong></p>
Correct Answer: A,B,D