A quadratic inequality asks not where a parabola crosses the horizontal axis, but where it lies above or below it. The roots of the corresponding quadratic equation mark the boundaries, the leading coefficient determines which way the parabola opens, and the discriminant decides whether there are two boundaries, one, or none. The solution is always readable from the shape of the parabola, and the sign chart formalizes that reading into a systematic procedure.
Definition and Standard Form
A quadratic inequality in one variable is any inequality that can be written as
ax2+bx+c>0
(or with <, ≤, or ≥), where a, b, c are real constants and a=0. The expression on the left is a quadratic polynomial, and the inequality asks where that polynomial takes positive or negative values.
Every quadratic inequality can be brought into standard form by moving all terms to one side. The inequality x2<3x+10 becomes x2−3x−10<0 after subtracting 3x+10 from both sides. The right-hand side is zero, and the question reduces to determining the sign of the left-hand side across the number line.
The solution depends entirely on the roots of the corresponding quadratic equationax2+bx+c=0. These roots, if they exist, are the boundary points — the values where the expression changes from positive to negative or vice versa. The sign of a determines which regions are positive and which are negative.
The Role of the Discriminant
The discriminant Δ=b2−4ac determines how many real roots the quadratic has, and therefore how many boundary points appear on the number line. This settles the structure of the solution before any interval testing begins.
When Δ>0, two distinct real roots r1<r2 exist. They divide the number line into three intervals: (−∞,r1), (r1,r2), and (r2,∞). The sign of the quadratic alternates across these intervals, and the solution is either the middle interval, the two outer intervals, or some combination depending on the inequality direction and the sign of a.
When Δ=0, one repeated root r exists. The quadratic factors as a(x−r)2, which is a perfect square scaled by a. The expression equals zero at x=r and has the sign of a everywhere else. The number line is effectively split into two intervals with identical sign, separated by a single point of zero.
When Δ<0, no real roots exist. The quadratic never touches zero and maintains one sign throughout the entire real line — the sign of a. The inequality is either satisfied everywhere or nowhere, with no intervals to analyze.
Solving When the Discriminant Is Positive
When Δ>0, the two roots r1<r2 create three intervals. The sign of the quadratic in these intervals follows a fixed pattern determined by the leading coefficient a.
If a>0, the parabola opens upward. The quadratic is positive on the outer intervals (−∞,r1) and (r2,∞), and negative on the inner interval (r1,r2). The expression dips below zero between the roots and rises above zero outside them.
If a<0, the parabola opens downward. The signs are reversed: the quadratic is negative on the outer intervals and positive on the inner interval (r1,r2). The expression rises above zero only between the roots.
To solve x2−5x+4>0, find the roots of x2−5x+4=0. Factoring gives (x−1)(x−4)=0, so r1=1 and r2=4. Since a=1>0, the expression is positive outside the roots. The solution is (−∞,1)∪(4,∞).
To solve x2−5x+4≤0 instead, select the interval where the expression is non-positive: [1,4]. The endpoints are included because the inequality is non-strict and the expression equals zero at the roots.
Solving When the Discriminant Is Zero
When Δ=0, the quadratic has a single repeated root r=−2ab, and the expression factors as a(x−r)2. The square (x−r)2 is non-negative for all real x and equals zero only at x=r. The sign of the entire expression is therefore determined by a alone, except at the single point x=r where the expression is zero.
If a>0: the expression is positive everywhere except at x=r, where it is zero. The inequality ax2+bx+c>0 is satisfied by all x=r, and the solution is (−∞,r)∪(r,∞). The non-strict version ax2+bx+c≥0 is satisfied by every real number: (−∞,∞).
If a<0: the expression is negative everywhere except at x=r, where it is zero. The inequality ax2+bx+c<0 is satisfied by all x=r, and the non-strict version ax2+bx+c≤0 holds for all real x. The inequality ax2+bx+c>0 has no solution — a negative times a non-negative square is never positive.
Consider x2−6x+9>0. The discriminant is 36−36=0, and the expression factors as (x−3)2. This is positive for all x=3 and zero at x=3. The solution is (−∞,3)∪(3,∞), the entire real line minus one point.
Solving When the Discriminant Is Negative
When Δ<0, the quadratic has no real roots. It never equals zero and never changes sign. The sign of the expression for every real x is the sign of the leading coefficient a.
If a>0: the expression is positive everywhere. The inequality ax2+bx+c>0 (or ≥0) is satisfied by every real number, and the solution is (−∞,∞). The inequality ax2+bx+c<0 (or ≤0) has no solution.
If a<0: the expression is negative everywhere. The inequality ax2+bx+c<0 (or ≤0) holds for all x, and ax2+bx+c>0 (or ≥0) has no solution.
The equation x2+x+1=0 has Δ=1−4=−3<0, and since a=1>0, the expression x2+x+1 is positive for every real x. The inequality x2+x+1>0 is universally true. The inequality x2+x+1<0 is universally false.
No sign chart is needed in this case. The absence of real roots means no critical points exist, so there is only one interval — the entire number line — and its sign is determined by a.
Sign Chart Method
The sign chart provides a systematic alternative to reasoning from the parabola's shape. It works by tracking the sign of each linear factor across the intervals created by the roots.
For the inequality (x−1)(x−4)>0, the roots are r1=1 and r2=4, creating three intervals. In the interval (−∞,1), both factors x−1 and x−4 are negative, so their product is positive. In (1,4), the factor x−1 is positive while x−4 is negative, so the product is negative. In (4,∞), both factors are positive, so the product is positive. The sign chart summarizes this:
(x−1)(x−4)productx<1−−+1<x<4+−−x>4+++
The solution to the >0 inequality consists of the intervals where the product is positive: (−∞,1)∪(4,∞).
When the quadratic does not factor neatly, the roots from the quadratic formula serve the same purpose. The sign chart does not require integer or rational roots — any pair of real numbers works as boundary points. The method fails only when no real roots exist (Δ<0), in which case the sign is constant and no chart is needed.
Connection to the Parabola
The inequality ax2+bx+c>0 asks a geometric question: where does the parabola y=ax2+bx+c lie above the x-axis? The inequality ax2+bx+c<0 asks where it lies below.
When a>0, the parabola opens upward. It dips below the axis between its roots (if they exist) and rises above the axis outside them. The >0 inequality selects the outer regions; the <0 inequality selects the inner region.
When a<0, the parabola opens downward. It rises above the axis between its roots and falls below outside. The >0 inequality selects the inner region; the <0 inequality selects the outer regions.
This geometric reading makes the sign chart unnecessary for many students — once the roots and the direction of opening are identified, the solution is visible from the shape of the curve. The sign chart is the algebraic encoding of the same information, and either approach yields identical results.
The vertex of the parabola, located at x=−2ab, is the point of maximum depth below the axis (when a>0) or maximum height above it (when a<0). Its position does not directly affect the solution set of the inequality, but it determines how far the parabola penetrates into the region of the opposite sign.
Reducible Inequalities
Certain inequalities of degree higher than two can be reduced to quadratic inequalities through substitution, following the same strategy used for reducible equations.
The bi-quadratic inequality x4−5x2+4>0 becomes u2−5u+4>0 with the substitution u=x2. The quadratic factors as (u−1)(u−4)>0, which is satisfied when u<1 or u>4. Back-substituting: x2<1 gives −1<x<1, and x2>4 gives x<−2 or x>2. The complete solution is (−∞,−2)∪(−1,1)∪(2,∞).
The back-substitution step requires care. The condition u=x2 forces u≥0, so any solution for u that is negative produces no real values of x. Additionally, each positive value of u splits into two symmetric values x=±u, potentially doubling the number of boundary points on the x-axis.
The substitution pattern extends to any inequality of the form a[f(x)]2+b[f(x)]+c>0. Setting u=f(x) reduces it to a quadratic inequality in u, and the solution in u is translated back through the function f — subject to whatever domain and range constraints f imposes.