The least integral value alpha of x such that | vedclass

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(D) Given the inequality: $\frac{x - 5}{(x + 7)(x - 2)} > 0$. We use the wavy curve method (sign scheme) to find the intervals. The critical points ar

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

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(D) Given the inequality: $\frac{x - 5}{(x + 7)(x - 2)} > 0$. We use the wavy curve method (sign scheme) to find the intervals. The critical points are $x = -7, 2, 5$. Testing the intervals: For $x > 5$,the expression is positive. For $2 For $-7 For $x The solution set is $x \in (-7, 2) \cup (5, \infty)$. The integral values in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$. The integral values 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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