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For b < 0, if x1 and x2 are the roots of 2x² + 6x + b = 0 and satisfy x1/x2 + x2/x1 < k, find the value of k.

1. -3
2. -5
3. -6
4. -2

Correct answer: -2

Solution

By Vieta's: x1+x2 = -6/2 = -3, x1*x2 = b/2. x1/x2 + x2/x1 = (x1²+x2²)/(x1x2) = ((x1+x2)² - 2x1x2)/(x1x2) = (9 - b)/(b/2) = 2(9-b)/b = 18/b - 2. For b<0: 18/b is negative, so 18/b - 2 < -2. The expression is always less than -2 for b<0. Therefore k = -2 is the least upper bound.

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