Step-by-Step Solution
Key Concept: Square both sides strategically and recognize that the cross term 2√[(x-2)(4-x)] becomes a constraint equation. The domain restriction 2 ≤ x ≤ 4 combined with the algebraic constraint uniquely determines x.
<p><strong>Step 1: Determine domain.</strong> For √(x-2) and √(4-x) to be real, we need x-2 ≥ 0 and 4-x ≥ 0, so <strong>2 ≤ x ≤ 4</strong>.</p><p><strong>Step 2: Square both sides.</strong> (√(x-2) + √(4-x))² = 2²<br>x-2 + 2√[(x-2)(4-x)] + 4-x = 4<br>2 + 2√[(x-2)(4-x)] = 4<br>2√[(x-2)(4-x)] = 2<br>√[(x-2)(4-x)] = 1</p><p><strong>Step 3: Square again.</strong> (x-2)(4-x) = 1<br>4x - x² - 8 + 2x = 1<br>-x² + 6x - 8 = 1<br>x² - 6x + 9 = 0<br>(x-3)² = 0<br><strong>x = 3</strong></p><p><strong>Step 4: Verify in original equation.</strong> √(3-2) + √(4-3) = √1 + √1 = 1 + 1 = 2 ✓</p><p>∴ <strong>Answer: x = 3</strong></p>
Correct Answer: x = 3