You finish solving x^2 - 4 > 0 and write x \in (-\infty, -2) \cup (2, \infty). You close your book. Question: is the answer right? You think so. The algebra looked clean. You flipped the sign where you were supposed to. No factor got dropped, as far as you remember.
Here is the cheapest, most reliable thirty seconds you can spend in an inequality problem: pick one number from each region of the real line and plug it into the original inequality. If the number you picked is inside your claimed solution, the inequality should hold. If it is outside your claimed solution, the inequality should fail. If either expectation is violated, something is wrong and you need to go back.
This is not double-checking the algebra — you already did that. This is testing the answer against the original question, which is a completely different check and catches completely different bugs.
The habit, in four steps
- Solve the inequality. Write the interval.
- Mark the critical points (the endpoints of your intervals) on an imaginary number line. These split \mathbb{R} into regions.
- Pick one convenient test value from each region — usually an easy integer like 0, \pm 1, \pm 3, or 10.
- Plug each test value into the original inequality (not a rearranged form). For values inside your claimed solution, the inequality should hold. For values outside, it should fail.
If any test disagrees with your answer, stop and find the bug before you move on. If all tests agree, the probability that your answer is wrong has just dropped dramatically.
Worked example: x^2 - 4 > 0
Suppose you solved this and got x \in (-\infty, -2) \cup (2, \infty). Let's sanity-check.
The critical points are -2 and 2. They split the real line into three regions: (-\infty, -2), (-2, 2), and (2, \infty). Pick one test value from each — the easiest choices are -3, 0, and 3.
Now plug in. Test x = -3 (inside the claimed region (-\infty, -2)):
Good — the inequality holds, as expected for a point in the solution. Test x = 0 (inside the claimed excluded region (-2, 2)):
Also good: the inequality fails, as expected for a point you excluded. Test x = 3 (inside the claimed region (2, \infty)):
All three tests agree with the claimed answer. The solution (-\infty, -2) \cup (2, \infty) passes the sanity check.
What a failed check actually looks like
Suppose a rushed student forgets the flip on a step and writes the "solution" to x^2 - 4 > 0 as x \in (-2, 2) — the opposite of the right answer. This is the kind of mistake that happens at 11pm when you are on your fourth problem and the sign-flip muscle memory is tired.
Plug x = 0, which is inside the claimed solution (-2, 2):
The test fails. A number your answer said should work, doesn't. That single failing test is a non-negotiable red flag: your answer is wrong, and you now know where to look — the step where the sign got flipped or the regions got swapped. Thirty seconds spent on a test value just saved you a wrong answer on the page.
Compare this to what happens without the sanity check: you move on, you lose the marks, and you don't even know there was a bug until the teacher returns the paper.
Why this catches bugs algebra can't
The sanity check is powerful because it is independent of the process that produced the answer. When you re-derive the solution by checking each step, you are using the same brain and often the same blind spots. If you got the sign flip wrong the first time, you will very likely get it wrong again on the re-derivation, because the mistake is built into how you see the problem.
Plugging a number is different. You are asking the original inequality a yes/no question at a specific point. The original inequality doesn't care what you think the answer is — it only cares whether the arithmetic works out. That is why the check is so reliable: the inequality itself is the arbiter.
The bugs this catches, in order of how often they actually happen on exam papers:
- Sign flip forgotten or applied when it shouldn't have been. A single sign error usually inverts an entire region. A test point in the flipped region fails immediately.
- A factor dropped during multiplication or factorisation. If you solved x(x - 3)(x + 1) > 0 by ignoring (x + 1), test values near x = -1 will expose the bug.
- Endpoints wrong — < vs \le confusion. Test the endpoint itself: does it satisfy the original inequality? If yes, the bracket should be square. If no, round.
- Whole-line vs empty-set errors. |x| < -3 has no solutions; |x| > -3 has all of \mathbb{R}. If you wrote one when you meant the other, a single test value will prove it.
How to pick good test values
Cheap is good. Boring is good. You want numbers that make the arithmetic trivial so you focus on the answer, not the calculation.
- Zero is almost always a great pick — if it lies in a region you haven't tested, use it first.
- Large positive and large negative values (like \pm 10 or \pm 100) reveal the behaviour of unbounded rays quickly.
- Avoid endpoints themselves unless you are specifically testing open-vs-closed. At a critical point the original inequality often evaluates to zero or an undefined form, which does not discriminate.
- For polynomial inequalities with several factors, pick one test value strictly between each pair of consecutive roots. The roots split the line into n + 1 regions for n distinct roots; you need one test from each.
When not to bother
If you are told to solve 2x > 6 and you write x > 3, you do not need to sanity-check — there is one step, and the mental cost of the plug-in is comparable to re-reading the line. Save the habit for problems where the algebra has at least two moves, where sign flips were involved, or where the answer is a union of intervals (those are the ones most prone to region-swapping).
Rule of thumb: if the answer is a union of more than one interval, or if any step involved dividing by a variable or multiplying by a negative, the sanity check is almost free insurance. Take it.
The connection to sign charts
For quadratic and polynomial inequalities, the sanity check is really a compressed version of the sign chart method: you mark the roots, and test one point in each region to decide the sign. When you use a sign chart to solve the inequality, you are already plugging a test number into each region — the sanity check is then automatic, because the chart is the check. The habit in this article is what you do when you solved the inequality some other way (say, by a pure algebraic manipulation) and want to verify independently.
For linear inequalities, a single test value from each side of the critical point is enough. For absolute-value inequalities that split into two rays, test one value from each ray and one from the gap in between. The pattern scales: one value per region, and you will sleep well at night.