The least integral value alpha of x such that | vedclass

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(D) Given the inequality: $\frac{x - 5}{x^2 + 5x - 14} > 0$.Factor the denominator: $x^2 + 5x - 14 = (x + 7)(x - 2)$.So,the inequality is $\frac{x - 5

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$\alpha^2 + 5\alpha - 6 = 0$

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(D) Given the inequality: $\frac{x - 5}{x^2 + 5x - 14} > 0$.Factor the denominator: $x^2 + 5x - 14 = (x + 7)(x - 2)$.So,the inequality is $\frac{x - 5}{(x + 7)(x - 2)} > 0$.Using the wavy curve method (sign scheme),the critical points are $x = -7, 2, 5$.The expression is positive in the intervals $(-7, 2) \cup (5, \infty)$.The integers in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.The integers in the interval $(5, \infty)$ are $\{6, 7, 8, \dots\}$.The least integral value $\alpha$ is $-6$.Checking the options for $\alpha = -6$:$(A)$ $(-6)^2 + 3(-6) - 4 = 36 - 18 - 4 = 14 
eq 0$.$(B)$ $(-6)^2 - 5(-6) + 4 = 36 + 30 + 4 = 70 
eq 0$.$(C)$ $(-6)^2 - 7(-6) + 6 = 36 + 42 + 6 = 84 
eq 0$.$(D)$ $(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0$.Thus,$\alpha = -6$ satisfies option $(D)$.

The least integral value $\alpha$ of $x$ such that $\frac{x - 5}{x^2 + 5x - 14} > 0$ satisfies:

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