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The perpendicular distance from point A(alpha, alpha²) to the line x + y + 3 = 0 is 2*sqrt(2). If point A lies in the first quadrant, how many values of alpha are possible?
- 0
- 1
- 2
- infinite
Correct answer: 1
Solution
|alpha + alpha² + 3| / sqrt(2) = 2*sqrt(2) => |alpha² + alpha + 3| = 4. Case 1: alpha² + alpha + 3 = 4 => alpha² + alpha - 1 = 0 => alpha = (-1 + sqrt(5))/2 (positive, 1st quadrant) or (-1-sqrt(5))/2 (negative). Case 2: alpha² + alpha + 3 = -4 => alpha² + alpha + 7 = 0; discriminant = 1-28 < 0, no real roots. Exactly one value alpha = (-1+sqrt(5))/2 ≈ 0.618 lies in the first quadrant.
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