Finding the Roots Is the Easy Half of Any Quadratic Inequality Calculator
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Marko Šinko
Co-Founder & Lead Developer, AI Math Calculator
A quadratic inequality calculator and a student solving x² − 5x + 6 > 0 by hand usually part company at the same moment. Both find x = 2 and x = 3 without trouble. The student then writes 2 < x < 3 and stops. The calculator returns (−∞, 2) ∪ (3, ∞) — the two pieces outside the roots, which is the exact opposite interval.
Nobody gets that wrong because they cannot factor. They get it wrong because factoring answers a different question. Roots tell you where the expression equals zero; an inequality asks where it is bigger or smaller than zero, and that is a question about the regions between the roots. Below is the two-step method that keeps the two apart, the sign chart that does the second step in under a minute, and the six outcomes you can end up with — including the two that have no roots at all.
Roots First, Then Ask Which Side of Zero You Want
Every quadratic inequality reduces to the same shape: get everything onto one side so the right-hand side is 0, then read the sign of what is left. That first move is not cosmetic. The sign of 2x² + 7x − 4 tells you something; the sign of 2x² + 7x compared against 4 tells you nothing, because you cannot compare a sign to a moving target.
Take 2x² + 7x ≥ 4. Subtract 4 from both sides and you have 2x² + 7x − 4 ≥ 0. Now factor: (2x − 1)(x + 4) ≥ 0, so the critical points are x = 0.5 and x = −4. If factoring stalls, the quadratic equation calculator will grind out the same two roots from the formula, and the discriminant calculator tells you in advance whether there will be two, one, or none.
Those two numbers are boundaries, not answers. They chop the number line into three regions, and inside each region the expression cannot change sign — a continuous function only flips from positive to negative by passing through zero, and the only places it hits zero are the roots you just found. That single fact is what makes the whole method work, and it is why one test value per region is enough.
Building the Sign Chart in Under a Minute
Draw a line, mark the critical points, and pick one convenient number from each region. Convenient means whatever is easiest to square in your head — for −4 and 0.5, try x = −5, x = 0 and x = 1. Substitute each into 2x² + 7x − 4:
| Region | Test x | 2x² + 7x − 4 | Sign | Keep for ≥ 0? |
|---|---|---|---|---|
| x < −4 | −5 | 50 − 35 − 4 = 11 | + | Yes |
| −4 < x < 0.5 | 0 | 0 + 0 − 4 = −4 | − | No |
| x > 0.5 | 1 | 2 + 7 − 4 = 5 | + | Yes |
Answer: (−∞, −4] ∪ [0.5, ∞). Three substitutions, no graphing, no guesswork. And notice the shortcut hiding in that table — for a genuine quadratic the signs always alternate as you cross each simple root, so once you know the sign of the far-right region you know all of them. The far-right region takes the sign of a, because for huge x the ax² term dwarfs everything else. That is a two-second check on any answer you produce.
Bracket or Parenthesis? The Rule That Decides Every Endpoint
Interval notation trips people up far more often than the algebra does. The rule is short: an endpoint gets a square bracket when the number itself satisfies the inequality, and a round one when it does not.
At a root, the expression equals exactly 0. Ask whether 0 satisfies your relation. For ≥ 0 and ≤ 0 it does, so the root belongs in the answer and earns a square bracket. For > 0 and < 0 it does not — zero is neither greater nor less than zero — so the root is excluded and gets a round bracket. Infinity is never a number you can reach, so ∞ and −∞ always take round brackets, in every single problem.
One more habit worth building: write the pieces left to right and join them with ∪, never with a comma. (−∞, 2) ∪ (3, ∞) is a union of two sets. Writing (−∞, 2), (3, ∞) or, worse, 3 < x < 2 turns a correct answer into an unreadable one. If unions and endpoint brackets are the part you keep losing marks on, the interval notation calculator converts between inequality form and interval form on its own, and compound inequalities follow exactly the same bracket rule.
Six Answers a Quadratic Inequality Calculator Can Return, and Two Have No Roots
Once you fix a > 0 (and you always can — multiply through by −1 and flip the sign), the discriminant decides the shape of the answer completely. There are only six possibilities, and three of them are the ones people never see coming:
| Discriminant | Relation | Solution shape | Example |
|---|---|---|---|
| D > 0 | > 0 | two outer rays | (−∞, 2) ∪ (3, ∞) |
| D > 0 | < 0 | one middle interval | (2, 3) |
| D = 0 | > 0 | everything except one point | (−∞, 3) ∪ (3, ∞) |
| D = 0 | ≤ 0 | a single point | {3} |
| D < 0 | > 0 | all real numbers | (−∞, ∞) |
| D < 0 | < 0 | no solution | ∅ |
The D = 0 row is the classic exam trap. Solve x² − 6x + 9 > 0 and the roots collapse to the single value x = 3. The parabola touches the axis there and bounces straight back up, so the expression is positive everywhere else — the answer is every real number except 3. Change that > to a ≥ and suddenly 3 qualifies too, giving all of ℝ. Change it to ≤ and only x = 3 survives. Three nearly identical questions, three completely different answers.
Three Inequalities That Behave Nothing Like Each Other
Irrational roots: x² − 2x − 4 < 0. This one will not factor. The discriminant is 4 + 16 = 20, so the roots are (2 ± √20)/2 = 1 ± √5, roughly −1.236 and 3.236. Since a is positive and we want the expression below zero, the answer is the middle piece: (1 − √5, 1 + √5). Leave it in surd form. Rounding to (−1.24, 3.24) quietly excludes numbers that genuinely satisfy the inequality.
Negative leading coefficient: −x² + 4 ≥ 0. The roots are ±2, but the parabola opens downward, so the positive part is the section between the roots, not outside them: [−2, 2]. If flipping your mental picture of the parabola feels risky, multiply through by −1 first — x² − 4 ≤ 0, remembering to reverse the inequality sign — and you are back to the upward-opening case you trust.
No real roots: x² + 2x + 5 > 0. Discriminant −16, so the parabola never touches the axis. Completing the square explains why: x² + 2x + 5 = (x + 1)² + 4, and a square plus 4 is at least 4 no matter what x does. The answer is all real numbers. Flip it to x² + 2x + 5 < 0 and the answer is the empty set — not zero, not the origin, but ∅, a set with nothing in it. Those two are worth recognising on sight, because a sign chart with no critical points looks broken until you realise there was only ever one region to test.
Why You Can Never Multiply an Inequality by x
Here is the mistake that survives long past school. Given 6/x < x, it is tempting to multiply both sides by x and solve 6 < x², giving x < −√6 or x > √6. Test x = −1: the original says 6/(−1) = −6 < −1, which is true — yet−1 is nowhere in that answer. The method dropped a whole chunk of the solution set.
Multiplying an inequality by a negative number reverses it, and x has no fixed sign, so you do not know whether to reverse or not. The fix is the same first step as always: move everything to one side. 6/x − x < 0 becomes (6 − x²)/x < 0, and now you build a sign chart with three critical points — x = ±√6 where the numerator vanishes, and x = 0 where the denominator does. Test each of the four regions and the correct answer, (−√6, 0) ∪ (√6, ∞), appears: the shortcut had wrongly claimed everything left of −√6 and wrongly thrown away everything between −√6 and 0.
The same warning applies to squaring both sides and to cancelling a factor of x, and it is why the sign-chart method is worth more than any shortcut. Dividing by a constant is safe as long as you flip the sign when that constant is negative — the familiar rule from linear inequalities — but a variable is never safe.
Cubics, Rational Expressions, and the Same Sign Chart
Nothing in the method depends on the highest power being 2. To solve any polynomial inequality, factor it fully, mark every root on the line, test one point per region, and collect the pieces you want. For x³ − 4x ≥ 0, factor to x(x − 2)(x + 2) ≥ 0, giving critical points −2, 0 and 2, four regions, and the answer [−2, 0] ∪ [2, ∞). The same routine handles anything the polynomial calculator can factor for you.
One extra rule covers rational inequalities: a value that makes the denominator zero is always a critical point, and it is always excluded, even when the relation is ≤ or ≥. It marks a place the expression is undefined, not a place it equals zero — which also means it belongs on your list of domain restrictions. Multiplicity is the other refinement: at a repeated root such as (x − 1)², the sign touches zero and comes back on the same side instead of crossing, exactly as the D = 0 quadratic did. Odd powers cross, even powers bounce.
Sign charts also do real work outside algebra class. Feed the derivative of a function into one and the positive regions are where the function increases; feed in the second derivative and the sign tells you which way the curve bends. If you want the formal treatment of why a continuous function cannot change sign without passing through zero, the intermediate value theorem is the result doing the heavy lifting behind every sign chart on this page.
