Quadratic Equations
Equations involving radicals
Grade 11

Question:

<p>The sum of the solutions of the equation \(\left|\sqrt{x} - 2\right| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0\), \((x > 0)\) is equal to __________.</p>

Step-by-Step Solution

Key Concept: Substitute y = √x to convert the absolute value equation into cases, then solve each case separately while checking which solutions satisfy the original domain and equation constraints.
<p><strong>Step 1:</strong> Let y = √x where y > 0. The equation becomes: |y - 2| + y(y - 4) + 2 = 0</p><p><strong>Step 2:</strong> Expand: |y - 2| + y² - 4y + 2 = 0, so |y - 2| = -y² + 4y - 2</p><p><strong>Step 3:</strong> <strong>Case 1:</strong> If y ≥ 2, then y - 2 = -y² + 4y - 2<br>⟹ y² - 3y = 0<br>⟹ y(y - 3) = 0<br>⟹ y = 3 (since y ≥ 2, y = 0 rejected)<br>Check: |3 - 2| + 9 - 12 + 2 = 1 - 1 = 0 ✓</p><p><strong>Step 4:</strong> <strong>Case 2:</strong> If y < 2, then -(y - 2) = -y² + 4y - 2<br>⟹ -y + 2 = -y² + 4y - 2<br>⟹ y² - 5y + 4 = 0<br>⟹ (y - 1)(y - 4) = 0<br>⟹ y = 1 or y = 4<br>Since y < 2, only y = 1 is valid.<br>Check: |1 - 2| + 1 - 4 + 2 = 1 - 1 = 0 ✓</p><p><strong>Step 5:</strong> From y = √x: when y = 1, x = 1; when y = 3, x = 9</p><p>∴ Answer: 1 + 9 = <strong>10</strong></p>
Correct Answer: 1

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