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The least integral value $$\alpha$$ of $$x$$ such that $$\frac{x-5}{x^2+5x-14} > 0$$, satisfies :
$$\frac{x-5}{x^2+5x-14} > 0$$
$$x^2 + 5x - 14 = (x + 7)(x - 2)$$
$$\frac{x-5}{(x+7)(x-2)} > 0$$
The solution set for the inequality is $$x \in (-7, 2) \cup (5, \infty)$$.
The smallest (least) integer in this combined set is $$\alpha = -6$$
Correctly satisfies option (D): $$\alpha^2 + 5\alpha - 6 = 0$$
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