<p>Find all possible values of \(\dfrac{x^2+1}{x^2-2}\).</p>
Step-by-Step Solution
Key Concept: Set y = (x² + 1)/(x² - 2) and rearrange to get a quadratic in x². For real x to exist, the discriminant must be non-negative, which constrains the possible values of y.
<p><strong>Step 1:</strong> Let y = (x² + 1)/(x² - 2). Rearrange to get a quadratic in x²:</p><p>y(x² - 2) = x² + 1</p><p>yx² - 2y = x² + 1</p><p>yx² - x² = 2y + 1</p><p>x²(y - 1) = 2y + 1</p><p>x² = (2y + 1)/(y - 1)</p><p><strong>Step 2:</strong> For real x to exist, we need x² ≥ 0 (and y ≠ 1):</p><p>(2y + 1)/(y - 1) ≥ 0</p><p><strong>Step 3:</strong> Analyze the sign of (2y + 1)/(y - 1):</p><p>• Numerator: 2y + 1 = 0 when y = -1/2</p><p>• Denominator: y - 1 = 0 when y = 1</p><p><strong>Step 4:</strong> Sign analysis on intervals:</p><p>• y < -1/2: negative/negative = positive ✓</p><p>• -1/2 < y < 1: positive/negative = negative ✗</p><p>• y > 1: positive/positive = positive ✓</p><p>• y = -1/2: equals 0 ✓</p><p><strong>Step 5:</strong> Note that y = 1 is excluded (makes denominator zero in original expression). Also verify x² ≠ 2 (which would make original denominator zero), but this is automatically handled.</p><p>∴ Answer: <strong>y ≤ -1/2 or y > 1</strong></p>
Correct Answer: y ≤ -1/2 or y > 1