Algebra · Eqora guide
Quadratic inequality homework: test the intervals
Roots mark boundaries, not the whole answer. Check signs, endpoints and repeated roots before writing your solution set.
You factor a quadratic correctly, find two roots and write those numbers as your homework answer. The algebra may be right, but the question is still unfinished. A quadratic inequality asks which inputs make a comparison true, usually entire intervals rather than two isolated points. In quadratic inequality homework, roots organize the search; the inequality symbol tells you which regions and boundary points to retain.
Take x² − x − 6 ≤ 0. Its roots are −2 and 3, yet 0 also satisfies the inequality because −6 ≤ 0. We will determine the complete set, explain why one test point can represent an interval, and examine cases where the usual sign pattern changes. If a step remains unclear after your attempt, use Eqora to investigate that transition, then finish a different problem independently.
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Start quadratic inequality homework with zero on one side
Work over the real numbers unless the question states a narrower domain. Move every term to one side while preserving the comparison: x² ≤ x + 6 becomes x² − x − 6 ≤ 0. Call that polynomial f(x). Now the question is about the sign of f, not the sign of x alone. A negative input can produce a positive polynomial value, so do not shade the negative half-line just because the symbol is ≤.
Factor f(x) = (x + 2)(x − 3). Expanding the product gives x² − x − 6, a quick check that both the middle and constant terms survived. Solve f(x) = 0 to locate the boundaries −2 and 3. West Texas A&M’s quadratic inequality lesson distinguishes locating boundaries from testing the resulting intervals. We use our own examples rather than copying its exercises; the roots are checkpoints, not automatically the entire solution.
Find the sign on every interval
The ordered boundaries divide the line into (−∞, −2), (−2, 3) and (3, ∞). For x < −2, both x + 2 and x − 3 are negative, so their product is positive. Between −2 and 3, the first factor is positive and the second negative, giving a negative product. Above 3, both factors are positive and the product is positive again. Read each factor before multiplying the signs.
Choose convenient representatives −3, 0 and 4. Direct substitution into the original polynomial gives 6, −6 and 6 respectively, agreeing with the factor signs. These values diagnose arithmetic errors, but a few successful substitutions alone do not establish a complete solution. The structural reason is that each linear factor keeps its sign until it passes through its own zero. Therefore the product keeps a constant nonzero sign within each of these root-free intervals.
Choose endpoints separately from the open regions
Our target f(x) ≤ 0 accepts negative values and zero. Retain the middle interval and include both roots because f(−2) = f(3) = 0. The answer is [−2, 3], equivalently −2 ≤ x ≤ 3. If the question instead used f(x) < 0, the same open region would qualify but the roots would not: (−2, 3). Brackets report inclusion; they do not determine the signs of neighboring values.
For f(x) > 0, retain the two outside regions: (−∞, −2) ∪ (3, ∞). The union means either region works; it does not require a number to lie in both. With f(x) ≥ 0, include the roots as well: (−∞, −2] ∪ [3, ∞). Infinity is not a real endpoint that can be included, so always use an open parenthesis there. Match interval notation to a sentence before submitting.
Do not divide away an unknown sign
Dividing (x + 2)(x − 3) ≤ 0 by x − 3 without cases is unsafe. That factor is negative below 3, positive above 3 and zero at 3. Division could reverse the inequality, preserve it or be undefined depending on the very input you are trying to classify. A sign chart avoids this circular shortcut and preserves the boundary x = 3 instead of silently discarding it.
A known negative constant is different. Consider −2x² + 2x + 12 ≥ 0. Dividing both sides by −2 reverses the symbol and yields x² − x − 6 ≤ 0, so the solution is again [−2, 3]. Alternatively, multiply the previous sign pattern by −2 and select nonnegative values. Both routes agree. Our guide to checking answers without an answer key encourages comparing genuinely different representations rather than repeating one fragile manipulation.
Check repeated roots and missing real roots
A squared factor does not change sign as it passes through zero. For (x − 1)² ≤ 0, every square is nonnegative, so only x = 1 qualifies. With (x − 1)² < 0 there is no real solution; with > 0 every real number except 1 qualifies; with ≥ 0 every real number qualifies. The two sides of a repeated root have the same sign. Do not alternate plus and minus mechanically at every marked point.
A quadratic can also have no real roots. For x² + 4 < 0, no real input works because x² + 4 is at least 4. The corresponding inequality x² + 4 > 0 holds for all real x. A negative discriminant does not mean every associated inequality has no solution: it means the quadratic equation has no real roots. The constant sign and the requested comparison still decide the answer. Completing the square can make that sign immediately visible.
Use Eqora to inspect one disputed choice
First make your own attempt: keep the original symbol, the zero-side rewrite, factorization, ordered boundaries and signs. If AI support is allowed, capture one complete legible problem and the few working lines that contain your doubt. Keep personal details out of the image. Our clear math photo guide explains why a cropped comparison symbol, lost exponent or missing minus can change the problem before the explanation even begins.
In Eqora, check that the captured task really says ≤ rather than < and includes the whole quadratic. Inspect the suggested steps against your working. Ask a precise question such as: “Why does the middle interval work for (x + 2)(x − 3) ≤ 0, and why are both roots included?” Request the signs of both factors at x = 0 and a separate endpoint check. Explain the response back without treating confidence as mathematical evidence.
If an explanation gives only the roots, alternates signs across a double root or drops a valid endpoint, do not copy it. Return to the original expression, the sign argument above or your teacher’s method. You can work through the examples on paper without a paid tool; that does not claim every app feature or unlimited usage is free. Close the response once the disputed step is clear and use the next problem to test what you retained.
Use exact boundaries and respect the stated domain
Not every boundary is an integer. For x² − 2 < 0, the roots are −√2 and √2, giving (−√2, √2). Keep the exact roots in the answer instead of replacing them by rounded decimals. A rough graph or calculator display can suggest the region, but cannot reliably decide a point extremely close to a boundary. Algebra and the strictness of the original comparison settle that question.
If the problem restricts x to nonnegative numbers, intersect your algebraic answer with [0, ∞). Our first answer [−2, 3] then becomes [0, 3]. If only integer inputs are allowed, list 0, 1, 2 and 3 instead. This restriction comes from the task, not from the quadratic formula. A correct real interval may still need a final domain check in a counting or measurement problem.
Finish a changed example with the response closed
Solve 2x² − 5x − 3 > 0 independently. Factor it as (2x + 1)(x − 3); expanding recovers the original polynomial. The ordered roots are −1/2 and 3. Use −1, 0 and 4 as representatives, obtaining polynomial values 4, −3 and 9. The signs are positive, negative and positive, so the strict inequality selects (−∞, −1/2) ∪ (3, ∞). Neither boundary belongs because both make the polynomial zero.
Check the factors’ signs across entire intervals and substitute the roots separately. Try a value you excluded as well as one you included; a check that examines only accepted values can miss an extra region. Finally say why the answer contains two intervals rather than two roots, why the endpoints are open and why infinity is not included. Being able to justify those decisions on changed coefficients is the learning outcome, not reproducing an app’s final line.
