Quadratic inequalities
Learning Objectives
- Define a quadratic inequality and its solution set.
- Solve \(ax^2 + bx + c \lessgtr 0\) by building the sign table and reading off the intervals.
- Handle the boundary correctly: include the roots for \(\leq, \geq\), exclude them for \(<, >\).
- Treat the degenerate cases \(\Delta \leq 0\), where the answer is \(\mathbb{R}\), a single point, or \(\varnothing\).
- Reduce a disguised inequality to standard form before solving.
From "where is it zero" to "where is it positive"
In Unit 1 of the linear course we solved first-degree inequalities like \(2x - 6 > 0\). A quadratic inequality is the same kind of question, one degree higher:
\[x^2 - x - 6 > 0, \qquad 3x^2 + 2 \leq 7x, \qquad -x^2 + 4 \geq 0.\]
Here is where the previous lesson pays off. Solving such an inequality is nothing more than asking where the quadratic function is positive (or negative, or zero) — which is exactly what the sign table records. The sign table is not a side-topic; it is the whole method. We now make that precise.
Quadratic inequality
A quadratic inequality in the unknown \(x\) is an inequality equivalent to one of the forms
\[ax^2 + bx + c > 0, \quad \geq 0, \quad < 0, \quad \leq 0,\]
with \(a, b, c \in \mathbb{R}\) and \(a \neq 0\). To solve it is to find its solution set: all real \(x\) for which the stated relation holds.
The relations come in two kinds. The strict ones (\(<, >\)) exclude the boundary, where the expression equals zero; the non-strict ones (\(\leq, \geq\)) include it. Keeping track of that single distinction is where most of the care lies.
The method, in three steps
Every quadratic inequality yields to the same routine:
Standard form. Move everything to one side, so the inequality compares \(f(x) = ax^2 + bx + c\) with \(0\).
Sign table. Compute \(\Delta\); if there are real roots, find them and build the sign table of \(f\) from the previous lesson.
Read off. Select the intervals whose sign matches the inequality — and decide, from strict versus non-strict, whether the roots themselves belong to the solution set.
The only genuinely new work compared with Lesson 1 is step 3's boundary decision; steps 1–2 are algebra and the sign table you already own.
A strict inequality
Solve \(x^2 - x - 6 > 0\).
It is already in standard form with \(f(x) = x^2 - x - 6\). From the previous lesson, \(f(x) = (x + 2)(x - 3)\), roots \(-2\) and \(3\), \(a = 1 > 0\), so \(f\) is positive outside \([-2, 3]\) and negative inside. We want \(f(x) > 0\) — the positive part:
Because the inequality is strict, the roots \(-2\) and \(3\) — where \(f\) is \(0\), not positive — are excluded. The solution set is
\[(-\infty, -2) \cup (3, \infty).\]
Had the inequality been \(x^2 - x - 6 \geq 0\), the same two rays would apply but now including the endpoints: \((-\infty, -2] \cup [3, \infty)\). One symbol, two bracket changes.
Bring it to standard form first
Solve \(3x^2 + 2 \leq 7x\).
Move everything left: \(3x^2 - 7x + 2 \leq 0\). Now \(f(x) = 3x^2 - 7x + 2\) with \(a = 3 > 0\) and \(\Delta = 49 - 24 = 25\); the roots are \(x = \dfrac{7 \pm 5}{6}\), that is \(2\) and \(\tfrac{1}{3}\). An upward parabola is negative between its roots, and the relation \(\leq\) includes the boundary, so
\[3x^2 + 2 \leq 7x \iff 3x^2 - 7x + 2 \leq 0 \iff x \in \left[\tfrac{1}{3},\, 2\right].\]
A frequent slip is to read the inequality before moving the \(7x\) across — the sign table only describes \(f(x)\) compared with \(0\), never \(f(x)\) compared with some other expression. Standard form is not optional.
When \(\Delta \leq 0\): the whole line, a point, or nothing
Without roots there is no sign change, and the answer is one of the extremes. Three quick cases:
\(x^2 + x + 1 > 0\). Here \(\Delta = -3 < 0\) and \(a > 0\), so \(f > 0\) everywhere: the solution is all of \(\mathbb{R}\).
\(x^2 + x + 1 < 0\). Same \(f\), always positive, so it is never negative: the solution set is \(\varnothing\).
\((x - 3)^2 \leq 0\). Here \(\Delta = 0\) and \(f(x) = (x-3)^2 \geq 0\) with equality only at \(x = 3\). A square is \(\leq 0\) exactly when it is \(0\), so the solution is the single point \(\{3\}\). (The strict \((x-3)^2 < 0\) would give \(\varnothing\).)
These are not exotic — a missed discriminant check is precisely how a solver ends up writing an interval where the true answer is "everything" or "nothing".
Connection to Computer Science
Software that plans motion — a robot arm, a drone, a delivery vehicle — often must keep a quantity within safe limits: a speed below a cap, a distance above a minimum. When the quantity varies quadratically in time, "stay safe" becomes a quadratic inequality, and its solution set is exactly the stretch of time during which the plan is allowed. The program solves the inequality — roots, then intervals — to compute the window in which it may act. The bracket-versus-parenthesis distinction matters here too: it is the difference between "up to and including the limit" and "strictly under it".
Why you cannot divide an inequality by \(x\)
Faced with \(x^2 > x\), it is tempting to "cancel an \(x\)" and conclude \(x > 1\). This is wrong twice over, and the reasons are worth naming.
First, dividing an inequality by \(x\) is illegal when the sign of \(x\) is unknown: dividing by a negative number reverses the inequality, and \(x\) might be negative. Second, dividing by \(x\) silently assumes \(x \neq 0\), discarding a value that may matter. The linear-inequality habit of dividing through does not survive contact with an unknown sign.
The correct move is the one this lesson teaches: bring everything to one side. \(x^2 > x \iff x^2 - x > 0 \iff x(x - 1) > 0\), a quadratic with roots \(0\) and \(1\) and \(a = 1 > 0\). Positive outside the roots, so the solution is \((-\infty, 0) \cup (1, \infty)\) — which includes negative numbers like \(x = -2\) (indeed \(4 > -2\)) that the naive "\(x > 1\)" throws away. The sign table sees the whole line at once, which is exactly why it cannot be fooled by the sign of a hidden factor.
Exercises
Exercise 1
Choose the solution set of each inequality.
Exercise 2
Consider \(2x^2 + 3 \leq 7x\). Rewrite it as \(2x^2 - 7x + 3 \leq 0\) and find the two roots, smaller first.
Exercise 3
Decide whether each statement about quadratic inequalities is true or false.
Exercise 4
Choose the solution set.
Exercise 5
The inequality \(x^2 - x - 6 \leq 0\) has solution set \([-2, 3]\) (roots \(-2, 3\), \(a > 0\)). For each value, decide whether it satisfies the inequality.
Summary
A quadratic inequality compares \(ax^2 + bx + c\) with \(0\) using \(<, \leq, >, \geq\); solving it means finding all \(x\) that satisfy the relation.
Method: (1) standard form — everything on one side; (2) sign table from the previous lesson; (3) read off the matching intervals.
Boundary: strict \(<, >\) exclude the roots (open intervals); non-strict \(\leq, \geq\) include them (closed). This is the step that most needs care.
Degenerate cases (\(\Delta \leq 0\)): with no sign change the answer is all of \(\mathbb{R}\), the empty set, or — when \(\Delta = 0\) and the relation admits equality — a single point.
Never divide an inequality by \(x\) (unknown sign): bring everything to one side and let the sign table decide. Next: two curves meeting — a line and a parabola.