Solve an inequality

The three answers that are not a number

A quadratic inequality does not always end with two numbers and an interval, and the three cases that are left are exactly the ones people get wrong. If the discriminant is negative the trinomial never touches the axis: with the matching sign it is always true, that is every real number works, and with the opposite sign it has no solution at all. If the discriminant is zero the trinomial touches the axis at a single point: with a strict greater than every number works except that one, and with a strict less than none of them works.

These are correct answers, not mistakes: an empty set is an answer, and so is «every real number». The page writes them out in words as well as in symbols, because in a written test «impossible» and «no solution» are not the same sentence, and the second one is the right one. If what you have in front of you has an equals sign instead of a greater or a less than, the page you need is Quadratic Equation Solver.

The sign chart, drawn

The sign of a product is the product of the signs, and that is why an inequality is solved by factorising and then looking at one factor at a time. The page draws the chart the way you draw it in an exercise book: one row per factor, the points that make it zero in columns, the line solid where that factor is positive and dashed where it is negative. The last row is the sign of the whole expression, with the plus and the minus interval by interval.

The rows of the factors that sit in the denominator are grey, because they count for the sign exactly like the others but their zeros stay out of the solutions. The columns of the critical points are in increasing order, and irrational roots sit in the right column even when their value can only be read rounded. If you first need to factorise a polynomial on its own, there is Polynomial Factoring.

The zeros of the denominator are left out, and the page says why

In a rational inequality the number that makes the denominator zero never enters the solutions, not even when the sign is the non strict one: the fraction does not exist there, it is not that it equals zero. It is the most frequent mistake with rational inequalities, and a round bracket on its own does not teach it, because that very bracket is also used next to infinity. That is why the page writes it as a sentence: that value has to be left out because it makes the denominator zero.

There is one case where books slip too: if the same factor appears above and below and cancels out, the point still stays out of the domain. In something like (x-1)(x+2)/(x-1) the value 1 does not work, even though after cancelling it can no longer be seen. Here the domain is collected while the inequality is being read, before any cancelling, so that point is not lost. If all you need is the value of a numeric expression step by step, the page is Solve an expression step by step.

How to type it, and what I do not do

You type it whole, with the sign in the middle: x^2 - 5x + 6 > 0, or (x-1)/(x+2) >= 0, or already factorised as (x-1)(x+3)(x-5) < 0. The times sign can be left out (2x, 3(x+1)), the power is the ^ sign and the signs can also be typed as <= and >=. Both sides can be full: moving everything to the left is my job, and that step is written down.

What I do not do is declared instead of being guessed: no absolute values, no unknown under a root or in the exponent, no logarithms, no letter parameters in the coefficients (the answer would depend on the sign of that parameter) and no systems, that is two inequalities together like -1 < x < 3. And if after factorising a piece of degree higher than two is left that does not split, I say so instead of answering at random. For equations with two unknowns put together in a system there is System of Equations Solver.