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The least integral value α of x such that (x−5)/(x²+5x−14) > 0, satisfies:

1. α² + 3α − 4 = 0
2. α² − 5α + 4 = 0
3. α² − 7α + 6 = 0
4. α² + 5α − 6 = 0

Correct answer: α² + 5α − 6 = 0

Solution

(x-5)/((x+7)(x-2)) > 0 holds on (-7,2) U (5, infinity). The least integer in this set is x=-6 (since -7 is excluded). Then alpha=-6 satisfies alpha^2+5*alpha-6 = 36-30-6 = 0, option 3, not the stored option 1.

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