Quadratic Equations
Nature of Roots
Grade None
Question:
<p>The equation \(\sqrt{3x^2 + x + 5} = x - 3\), where \(x\) is real, has</p>
<p>no solution.</p>
<p>exactly one solution.</p>
<p>exactly two solutions.</p>
<p>exactly four solutions.</p>
Step-by-Step Solution
Key Concept: For √f(x) = g(x) to have real solutions, both sides must be non-negative AND their squares must be equal. This requires g(x) ≥ 0 as a necessary condition before squaring.
<p><strong>Step 1:</strong> For √(3x² + x + 5) = x - 3 to have real solutions, the RHS must be non-negative:</p><p>x - 3 ≥ 0 ⟹ x ≥ 3</p><p><strong>Step 2:</strong> Square both sides (valid since both sides are non-negative when x ≥ 3):</p><p>3x² + x + 5 = (x - 3)²</p><p>3x² + x + 5 = x² - 6x + 9</p><p>2x² + 7x - 4 = 0</p><p><strong>Step 3:</strong> Solve using the quadratic formula:</p><p>x = [-7 ± √(49 + 32)]/4 = [-7 ± √81]/4 = [-7 ± 9]/4</p><p>x = 1/2 or x = -4</p><p><strong>Step 4:</strong> Check against domain restriction x ≥ 3:</p><p>• x = 1/2: Does not satisfy x ≥ 3 ✗</p><p>• x = -4: Does not satisfy x ≥ 3 ✗</p><p><strong>Step 5:</strong> Verify by substitution (e.g., x = 1/2): √(3/4 + 1/2 + 5) = √(27/4) ≠ -5/2 (since RHS is negative)</p><p>∴ The equation has <strong>no real solutions</strong></p>
Correct Answer: A