College Algebra Quiz: Rational Inequalities Using Sign Charts

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Rational Inequalities Using Sign ChartsQuestion 1 of 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 $b$ ?

a = 1, b = -2

a = -1, b = 2

a = 1, b = 2

a = -1, b = -2

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College Algebra Quiz

College Algebra Quiz: Rational Inequalities Using Sign Charts

Practice Rational Inequalities Using Sign Charts in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Inequalities Using Sign Charts, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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 $b$ ?

$a = 1, b = -2$
$a = -1, b = 2$
$a = 1, b = 2$
$a = -1, b = -2$ (correct answer)

Explanation: 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.

Question 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?

Negative sign; the interval $(-3, 2)$ is part of the solution set
Positive sign; the interval $(-3, 2)$ is not part of the solution set
Negative sign; the interval $(-3, 2)$ is not part of the solution set (correct answer)
Positive sign; the interval $(-3, 2)$ is part of the solution set

Explanation: 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.

Question 3

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?

Correct solution: $x \leq 4$ ; No error was made in the student's work
Correct solution: $(-3, 4]$ ; Student forgot to exclude $x = -3$
Correct solution: $(-3, 4]$ ; Student failed to consider that $x + 3$ could be negative (correct answer)
Correct solution: $[-3, 4]$ ; Student should have used brackets instead of parentheses

Explanation: 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.

Question 4

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?

Zeros at $x = -2, 7$ ; vertical asymptotes at $x = 1, 4$ (correct answer)
Zeros at $x = -2, 1$ ; vertical asymptotes at $x = 4, 7$
Zeros at $x = 1, 7$ ; vertical asymptotes at $x = -2, 4$
Zeros at $x = -2, 4$ ; vertical asymptotes at $x = 1, 7$

Explanation: 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.

Question 5

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?

$x = 2$ only
$x = -2$ only (correct answer)
Both $x = 2$ and $x = -2$
Neither; all critical values create vertical asymptotes

Explanation: 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.