Least integral value alpha of x such that

Question

Question:

My solution:

According to me -6 is least possible integer which satisfies option d.

Textbook's solution

I don't understand what I'm doing wrong. Thank you for your help.

Answer 1

So we are given the inequality:

$$ \frac{\left(x-5\right)}{\left(x+7\right)\left(x-2\right)}>0 $$

Applying wavy-curve like you did or just looking at the graph of the inequality, we get our solution interval as:

$$ x\in (-7,2) \cup (5, \infty) $$

I have verified this answer over WolframAlpha:

Hence the least integral value that $x$ can take would be $-6$, followed by $-5,-4,-3...1$

I think we can see that $-6$ satisfies option D as :

$$ f(x) = x^2+5x-6, \ \ f(-6) = 0$$

So I am guessing that the correct answer should be $-6$.

Answer 2

Multiplying both sides by the square of the denominator on LHS gives $$(x-5)(x^2+5x-14)>0,$$ which factors as $$(x-5)(x+7)(x+2)>0.$$

Thus the values of $x$ satisfying the inequality are $-7<x<2$ or $x>5.$

So we can see that the least integer satisfying the inequality is $-6.$