Nonlinear inequalities are solved by finding the critical points where an expression equals zero or is undefined, then testing the intervals between those points to determine where the inequality holds true. This method works for quadratic, higher-degree polynomial, and rational inequalities alike. The core idea is the same every time: get zero on one side, find the critical points, build a sign chart, and read off the solution.
The General Process
Every nonlinear inequality follows the same four-step framework, regardless of the type of expression involved:
- Rewrite the inequality so one side is zero. Move all terms to one side so you’re comparing some expression f(x) to 0. This is essential because the method depends on tracking where f(x) is positive and where it’s negative.
- Find the critical points. These are the x-values where f(x) equals zero (its roots) and, for rational expressions, where f(x) is undefined. Factor the expression or use the quadratic formula to locate them.
- Build a sign chart. Plot the critical points on a number line. They divide the line into intervals. Pick any convenient test value from each interval, plug it into f(x), and record whether the result is positive or negative.
- Read off the solution. Select the intervals whose signs satisfy your original inequality. Use interval notation to write the answer, paying attention to whether endpoints are included.
The reason this works is that polynomials and rational expressions are continuous between their critical points. The only places where an expression can switch from positive to negative (or vice versa) are at its zeros or where it’s undefined. So once you know the sign in any one interval, it stays that sign throughout the entire interval.
Solving Quadratic Inequalities
Quadratic inequalities are the most common starting point. Suppose you need to solve x² − x − 6 < 0. First, factor the left side: (x − 3)(x + 2) < 0. The expression equals zero when x = 3 and x = −2. These two critical points split the number line into three intervals: (−∞, −2), (−2, 3), and (3, ∞).
Now pick a test point in each interval. For (−∞, −2), try x = −3: (−3 − 3)(−3 + 2) = (−6)(−1) = 6, which is positive. For (−2, 3), try x = 0: (0 − 3)(0 + 2) = (−3)(2) = −6, which is negative. For (3, ∞), try x = 4: (4 − 3)(4 + 2) = (1)(6) = 6, which is positive.
You need the expression to be less than zero, so the solution is the interval where the sign is negative: (−2, 3). Since the inequality is strict (<, not ≤), the endpoints are excluded and you use parentheses. If the inequality were ≤, you’d include the zeros and write [−2, 3].
When a quadratic doesn’t factor neatly, use the quadratic formula to find the zeros. The process after that is identical.
Higher-Degree Polynomial Inequalities
The same method scales to polynomials of any degree. Consider x⁴ + 4x³ − 12x² ≤ 0. Factor out x² first to get x²(x² + 4x − 12), then factor the remaining quadratic: x²(x + 6)(x − 2) ≤ 0. The zeros are x = −6, x = 0, and x = 2, creating four intervals to test.
Here’s where a critical warning applies: do not assume the signs alternate between intervals. With quadratics, the pattern positive-negative-positive (or its reverse) always holds because you have exactly two distinct roots. With higher-degree polynomials, repeated roots can cause the expression to touch zero without changing sign. In the example above, x = 0 is a repeated root (from the x² factor), so the expression stays non-negative on both sides of zero rather than flipping. You must test every interval individually.
After testing, suppose the sign chart reads: positive on (−∞, −6), negative on (−6, 0), positive on (0, 2), and negative on… you’d actually need to test to confirm, but the point is that plugging in real numbers is the only reliable way to determine each interval’s sign. Once you have the chart, select the intervals where the expression is negative or zero (since the inequality is ≤), and include the zeros themselves as endpoints with brackets.
Rational Inequalities
Rational inequalities, those involving fractions with variables in the denominator, follow the same framework with one extra rule: values that make the denominator zero are critical points, but they can never be part of the solution. They create vertical asymptotes or holes, not actual values of the function.
Take (x − 1)/(x + 3) ≥ 0 as an example. The numerator is zero when x = 1, making it a critical point. The denominator is zero when x = −3, making that a critical point too. You now have three intervals: (−∞, −3), (−3, 1), and (1, ∞).
Test each interval. For x = −4: (−4 − 1)/(−4 + 3) = (−5)/(−1) = 5, positive. For x = 0: (0 − 1)/(0 + 3) = −1/3, negative. For x = 2: (2 − 1)/(2 + 3) = 1/5, positive. The expression is non-negative on (−∞, −3) and (1, ∞). Since the inequality is ≥, you include x = 1 (where the expression equals zero) with a bracket. But x = −3 makes the denominator zero, so it’s always excluded with a parenthesis. The solution is (−∞, −3) ∪ [1, ∞).
One tempting shortcut with rational inequalities is to multiply both sides by the denominator to “clear the fraction.” This is dangerous because the denominator might be positive for some values of x and negative for others, and multiplying by a negative number flips the direction of the inequality. Unless you split the problem into separate cases based on the sign of the denominator, you’ll get wrong answers. The sign chart method avoids this problem entirely.
Writing Solutions in Interval Notation
Interval notation uses brackets and parentheses to communicate exactly which values are included. Parentheses mean “up to but not including” that endpoint. Brackets mean “including” that endpoint. Infinity always gets a parenthesis because it’s not a number you can reach.
A few quick examples: all numbers between −2 and 6, including −2 but not 6, is written [−2, 6). All numbers greater than 3 is (3, ∞). All numbers less than or equal to 5 is (−∞, 5].
When your solution has two or more separate intervals, join them with the union symbol ∪. For instance, if the solution is everything less than −3 or greater than or equal to 1, you’d write (−∞, −3) ∪ [1, ∞). The union symbol simply means “or,” combining the two pieces into one solution set.
Deciding When to Use Brackets vs. Parentheses
The choice between brackets and parentheses at each critical point depends on two things: the type of inequality and the type of critical point.
If the inequality is strict (< or >), use parentheses at every critical point. The expression at those points equals zero, and zero is neither greater than nor less than zero.
If the inequality is non-strict (≤ or ≥), use brackets at zeros of the expression because the expression equals zero there, and zero satisfies ≤ 0 or ≥ 0. However, for rational inequalities, any critical point where the expression is undefined (denominator equals zero) always gets a parenthesis, regardless of the inequality type. You can’t include a value in the solution if the expression doesn’t exist there.
Common Mistakes to Avoid
The most frequent error is assuming that signs alternate between intervals. This holds for simple quadratics with two distinct roots, but it breaks down with repeated roots or higher-degree polynomials. A factor like x² touches zero at x = 0 without changing sign, so intervals on both sides of that root can have the same sign. Always test every interval.
Another common mistake is forgetting to move everything to one side before starting. If you try to find critical points from something like x² − 3x > 2x + 6 without first rewriting it as x² − 5x − 6 > 0, you’ll identify the wrong critical points and get a wrong answer.
Finally, watch for domain restrictions you might overlook. Square roots, logarithms, and other functions impose their own domain limits beyond just the zeros and undefined points of a rational expression. If your inequality involves something like √(x − 1), the expression only exists when x ≥ 1, so your number line starts there.