Rational Inequalities Using Sign Charts - College Algebra

1. Consider the rational inequality $\frac{x^2 - 4}{x^2 + x - 2} < 0$. After factoring completely, which critical values create removable discontinuities (holes) rather than vertical asymptotes?
   Answer: $x = -2$ only Factor the numerator: $x^2 - 4 = (x-2)(x+2)$. Factor the denominator: $x^2 + x - 2 = (x+2)(x-1)$. So the inequality becomes $\frac{(x-2)(x+2)}{(x+2)(x-1)} < 0$. The factor $(x+2)$ appears in both numerator and denominator, creating a removable discontinuity (hole) at $x = -2$. After canceling the common factor: $\frac{x-2}{x-1} < 0$ for $x \neq -2$. The remaining critical values are $x = 1$ (vertical asymptote from denominator) and $x = 2$ (zero from numerator), but only $x = -2$ creates a removable discontinuity.
2. When solving $\frac{x^2 - 5x + 6}{x^2 - x - 12} > 0$, a student factors and gets $\frac{(x-2)(x-3)}{(x-4)(x+3)} > 0$. If the student tests the interval $(-3, 2)$ by substituting $x = 0$, what sign should they get, and what does this tell us about the solution?
   Answer: Negative sign; the interval $(-3, 2)$ is not part of the solution set Substituting x = 0 into $\frac{(x-2)(x-3)}{(x-4)(x+3)}$: numerator = (-2)(-3) = 6 > 0, denominator = (-4)(3) = -12 < 0. So the expression equals $\frac{6}{-12} = -\frac{1}{2} < 0$, giving a negative sign. Since we want the expression > 0 (positive), and the test point gives negative, the interval (-3, 2) is NOT part of the solution set. Choice A incorrectly concludes that negative means it's part of the solution. Choices B and D incorrectly calculate the sign as positive.
3. A rational inequality $\frac{f(x)}{g(x)} \leq 0$ has critical values at $x = -2, 1, 4, 7$. If the solution set is $[-2, 1] \cup (4, 7)$, which statement about the zeros and vertical asymptotes is correct?
   Answer: Zeros at $x = -2, 7$; vertical asymptotes at $x = 1, 4$ In the solution set [-2, 1] ∪ (4, 7), the endpoints that are included (x = -2, 1) correspond to zeros of the numerator where the rational expression equals zero. The endpoints that are excluded from intervals where the expression changes sign (x = 4, 7) correspond to vertical asymptotes where the denominator equals zero. Specifically: x = -2 is included, so it's a zero; x = 1 is included at the right endpoint, so it's a zero; x = 4 is excluded (parenthesis), so it's a vertical asymptote; x = 7 is excluded from the interval, so it's a vertical asymptote. Choice B incorrectly identifies x = 7 as a zero when it's excluded. Choice C incorrectly identifies x = -2 as a vertical asymptote when it's included. Choice D incorrectly identifies x = 4 as a zero when it's excluded.
4. A student incorrectly solves $\frac{2x-1}{x+3} \leq 1$ by cross-multiplying to get $2x - 1 \leq x + 3$, leading to $x \leq 4$. What is the correct solution set, and what error did the student make?
   Answer: Correct solution: $(-3, 4]$; Student failed to consider that $x + 3$ could be negative The student's error is cross-multiplying without considering the sign of $(x+3)$. When cross-multiplying an inequality, you must consider whether you're multiplying by a positive or negative quantity, which can flip the inequality sign. The correct approach: $\frac{2x-1}{x+3} \leq 1$ becomes $\frac{2x-1}{x+3} - 1 \leq 0$, which gives $\frac{2x-1-(x+3)}{x+3} = \frac{x-4}{x+3} \leq 0$. Using a sign chart with critical values $x = -3, 4$: for $x < -3$, both factors are negative so the quotient is positive; for $-3 < x < 4$, the numerator is negative and denominator is positive, so the quotient is negative; for $x > 4$, both factors are positive so the quotient is positive. We want $\leq 0$, so the solution is $(-3, 4]$, including $x = 4$ where the expression equals zero but excluding $x = -3$ where the expression is undefined.
5. The solution to $\frac{x^2 + ax + b}{x^2 - 9} > 0$ is $(-\infty, -3) \cup (-1, 2) \cup (3, \infty)$. What are the values of $a$ and $b$?
   Answer: $a = -1, b = -2$ When you encounter a rational inequality with a given solution set, you need to work backwards to find the original expression. The key insight is that the solution intervals tell you exactly where the expression changes sign. The denominator $x^2 - 9 = (x-3)(x+3)$ creates vertical asymptotes at $x = -3$ and $x = 3$, which matches the boundaries in our solution set. Since the inequality is positive on $(-\infty, -3) \cup (-1, 2) \cup (3, \infty)$, the expression must change sign at $x = -1$ and $x = 2$. This means the numerator $x^2 + ax + b$ must have zeros at these points. If the numerator equals zero when $x = -1$ and $x = 2$, then $x^2 + ax + b = (x+1)(x-2) = x^2 - x - 2$. Therefore, $a = -1$ and $b = -2$. Let's verify: $\frac{(x+1)(x-2)}{(x+3)(x-3)} > 0$. Testing the sign in each interval confirms this matches the given solution. Looking at the wrong answers: Choice A gives $a = 1, b = -2$, which would make the numerator $(x+2)(x-1)$, shifting the zeros to the wrong locations. Choice B gives $a = -1, b = 2$, resulting in a numerator with no real zeros since the discriminant is negative. Choice C gives $a = 1, b = 2$, which also produces no real zeros. Strategy tip: When working with rational inequalities, always identify the zeros and asymptotes first—they're the critical points where the expression changes sign and determine the solution intervals.

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