<p><strong>23.</strong> Suppose that \(x_1\) and \(x_2\) are the positive real solution of \(x^2 - bx + c = 0\) provided that \(x_1^2 + \sqrt{x_2^2 - 2x_2} = 2x_1 - 1\). The minimum value of \((b + c)\), is:</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas (b = x₁ + x₂, c = x₁x₂) combined with the constraint equation to express b + c in terms of one variable, then apply calculus or algebraic manipulation to find the minimum.
<p><strong>Step 1:</strong> From the given equation: x₁² + √(x₂² - 2x₂) = 2x₁ - 1</p><p>Rearrange: √(x₂² - 2x₂) = 2x₁ - 1 - x₁²</p><p><strong>Step 2:</strong> For the square root to be defined and real: x₂² - 2x₂ ≥ 0 ⟹ x₂(x₂ - 2) ≥ 0. Since x₂ > 0, we need x₂ ≥ 2.</p><p><strong>Step 3:</strong> For the RHS to be non-negative: 2x₁ - 1 - x₁² ≥ 0 ⟹ x₁² - 2x₁ + 1 ≤ 0 ⟹ (x₁ - 1)² ≤ 0</p><p>This forces x₁ = 1.</p><p><strong>Step 4:</strong> Substitute x₁ = 1 into the constraint: 1 + √(x₂² - 2x₂) = 2(1) - 1 = 1</p><p>Therefore: √(x₂² - 2x₂) = 0 ⟹ x₂² - 2x₂ = 0 ⟹ x₂ = 2 (since x₂ > 0)</p><p><strong>Step 5:</strong> By Vieta's formulas: b = x₁ + x₂ = 1 + 2 = 3, and c = x₁x₂ = 1(2) = 2</p><p><strong>Step 6:</strong> Therefore, b + c = 3 + 2 = 5</p><p>∴ Answer: B (minimum value of b + c is <strong>5</strong>)</p>
Correct Answer: B