Graphing Quadratic Inequalities Calculator

Enter your quadratic coefficients and choose the inequality direction to instantly find the solution set and interval notation.

📉 Quadratic Inequalities Calculator
Inequality: ax² + bx + c ◻ 0
Coefficient a (must be ≠ 0)
Coefficient b
x
Coefficient c
Inequality Direction
Solution Set
Interval Notation
Boundary Roots
Discriminant (D)
Parabola Opens
Inequality
Root Nature

📉 What is a Quadratic Inequality?

A quadratic inequality is a mathematical statement that compares a quadratic expression to zero using an inequality sign. The four standard forms are ax² + bx + c > 0, ax² + bx + c < 0, ax² + bx + c ≥ 0, and ax² + bx + c ≤ 0, where a ≠ 0. Unlike a quadratic equation (which finds specific points where the parabola crosses the x-axis), a quadratic inequality asks which values of x make the expression positive, negative, non-negative, or non-positive. The answer is typically a range of values, not a single number.

Quadratic inequalities appear across engineering, physics, economics, and computer science. In projectile motion, finding the time window during which an object is above a certain height reduces to solving h(t) > k, which is a quadratic inequality in t. In business, finding the range of production volumes that yield a positive profit can produce a quadratic inequality in units produced. In signal processing, analyzing the frequency ranges where a quadratic filter's gain exceeds a threshold uses the same technique.

The solution method relies on two key facts: first, the quadratic formula or discriminant tells you where (if anywhere) the expression equals zero, giving the boundary or critical points. Second, because a parabola is a smooth continuous curve, it can only change sign at these boundary points. Between any two consecutive boundary points, the expression maintains a constant sign. This means you only need to test one point in each region to know the sign of the entire region.

This calculator handles all three discriminant cases automatically. When D > 0 (two distinct real roots), it identifies the two critical points and expresses the solution as a union of intervals or a bounded interval depending on the parabola direction and inequality type. When D = 0 (one repeated root), it handles the special cases where the solution may be a single point or all reals except one point. When D < 0 (no real roots), the parabola never crosses the x-axis, so the solution is either all real numbers or the empty set.

📐 Formula

ax² + bx + c  ◻  0   where ◻ is >, ≥, <, or ≤
a = quadratic coefficient (must be non-zero); determines parabola direction
b = linear coefficient
c = constant term
D = b² − 4ac (discriminant; determines number of real roots)
Boundary roots: x  =  (−b ± √D) ÷ (2a)   when D ≥ 0
Roots x&sub1; and x&sub2; (with x&sub1; ≤ x&sub2;) are the critical points that divide the number line into regions.
D > 0: Two distinct real roots. Solution depends on sign of a and inequality direction.
D = 0: One repeated root. Solution may be a single point, all reals, or all reals except the vertex.
D < 0: No real roots. Solution is either all reals or no solution (empty set).
Solution rules for a > 0 (parabola opens upward)
ax² + bx + c > 0: x < x&sub1; or x > x&sub2;   (outside the roots)
ax² + bx + c < 0: x&sub1; < x < x&sub2;   (between the roots)
ax² + bx + c ≥ 0: x ≤ x&sub1; or x ≥ x&sub2;
ax² + bx + c ≤ 0: x&sub1; ≤ x ≤ x&sub2;
For a < 0 (opens downward), the inside/outside regions swap.

📖 How to Use This Calculator

Steps to Solve a Quadratic Inequality

1
Enter the coefficients a, b, and c for your quadratic inequality ax² + bx + c. Coefficient a must not be zero.
2
Choose the inequality direction from the dropdown: > 0 (greater than zero), ≥ 0, < 0, or ≤ 0.
3
Click Solve to see the discriminant, boundary roots, full solution set in plain language, and interval notation ready to copy.

💡 Example Calculations

Example 1 — Classic Upward Parabola, Positive Region

Solve x² − 5x + 6 > 0

1
Coefficients: a = 1, b = −5, c = 6. Discriminant: D = (−5)² − 4(1)(6) = 25 − 24 = 1
2
Roots: x = (5 ± 1) ÷ 2 → x&sub1; = 2, x&sub2; = 3. Parabola opens upward (a = 1 > 0).
3
For > 0 with upward parabola: solution is outside the roots.
Solution: x < 2 or x > 3  ·  Interval: (−∞, 2) ∪ (3, +∞)
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Example 2 — Upward Parabola, Negative Region

Solve x² − 5x + 6 < 0

1
Same roots as Example 1: x&sub1; = 2, x&sub2; = 3. Parabola opens upward.
2
For < 0 with upward parabola: solution is between the roots.
Solution: 2 < x < 3  ·  Interval: (2, 3)
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Example 3 — Downward Parabola

Solve −x² + 4x − 3 ≥ 0

1
Coefficients: a = −1, b = 4, c = −3. D = 16 − 12 = 4. Roots: x = (−4 ± 2) / (−2) → x&sub1; = 1, x&sub2; = 3.
2
Parabola opens downward (a = −1 < 0). For ≥ 0 with downward parabola: solution is between the roots (inclusive).
Solution: 1 ≤ x ≤ 3  ·  Interval: [1, 3]
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Example 4 — No Real Roots (All Reals)

Solve x² + x + 2 > 0

1
Coefficients: a = 1, b = 1, c = 2. D = 1 − 8 = −7 < 0. No real roots.
2
Parabola opens upward and never crosses x-axis, so x² + x + 2 is always positive.
Solution: All real numbers  ·  Interval: (−∞, +∞)
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❓ Frequently Asked Questions

How do you solve a quadratic inequality step by step?+
Step 1: Rearrange to ax² + bx + c > 0 (or <, >=, <=) with zero on the right side. Step 2: Solve the equality ax² + bx + c = 0 using the quadratic formula to find the boundary roots. Step 3: Determine the parabola direction from the sign of a. Step 4: Apply the rules: if a > 0 and inequality is > 0, solution is outside the roots; if a > 0 and inequality is < 0, solution is between the roots. Reverse for a < 0.
What is interval notation for a quadratic inequality solution?+
Interval notation uses ( ) for strict endpoints (excluded) and [ ] for non-strict endpoints (included). For x² - 5x + 6 > 0 with roots 2 and 3, the solution x < 2 or x > 3 is written (-∞, 2) ∪ (3, +∞). For the non-strict version >= 0, it becomes (-∞, 2] ∪ [3, +∞). The ∪ symbol denotes the union of two intervals.
What happens when the discriminant is negative in a quadratic inequality?+
When D < 0, the parabola has no real roots and never crosses the x-axis. If a > 0 (opens upward), the expression is always positive: ax² + bx + c > 0 for all real x (solution = all reals), and < 0 has no solution. If a < 0 (opens downward), the expression is always negative: < 0 for all real x, and > 0 has no solution. Non-strict versions (>= or <=) follow the same pattern.
What does it mean when a quadratic inequality has no solution?+
No solution means no real x satisfies the inequality. Example: x² + 4 < 0. Since x² + 4 >= 4 for all real x (minimum is 4 at x = 0), it is never negative. The solution set is the empty set, written ∅. This occurs when the parabola lies entirely on the wrong side of the x-axis for the given inequality direction.
How does the direction of the parabola affect the solution?+
When a > 0 the parabola opens upward: the quadratic is positive outside the roots and negative between them. When a < 0 it opens downward: positive between the roots and negative outside them. This means the same two roots produce completely different solution sets depending on whether a is positive or negative. For example, x² - 4 > 0 (a=1, roots ±2) gives x < -2 or x > 2, while -x² + 4 > 0 gives -2 < x < 2.
What is the difference between strict and non-strict quadratic inequalities?+
Strict inequalities (> or <) exclude the boundary roots from the solution. Non-strict (>= or <=) include them. At the boundary roots, ax² + bx + c = 0 exactly, so for > 0 the roots are not part of the solution; for >= 0 they are. In interval notation, strict uses parentheses () at boundary points and non-strict uses brackets []. The interior of the solution set is identical in both cases.
How do you graph a quadratic inequality on a number line?+
Mark the roots as boundary points on the number line (open circles for strict inequalities, filled circles for non-strict). Test a value in each region by substituting into the original inequality. Shade the regions where the inequality holds. For x² - 5x + 6 < 0 with roots 2 and 3: test x=0 (positive, not shaded), test x=2.5 (negative, shaded), test x=4 (positive, not shaded). Shade only the region between 2 and 3.
What happens when the quadratic has a repeated root?+
With a repeated root x₀ (D = 0, parabola tangent to x-axis), the parabola touches but does not cross the x-axis. For a > 0 and > 0: solution is all reals except x₀, written (-∞, x₀) ∪ (x₀, +∞). For a > 0 and >= 0: all real numbers. For a > 0 and < 0: no solution. For a > 0 and <= 0: only x = x₀. For a < 0, the inside/outside logic reverses.
How do you solve x squared minus 4 is greater than 0?+
Equation: x² - 4 = 0 gives x = ±2. Since a = 1 > 0 (parabola opens up) and we want > 0 (positive region outside roots): solution is x < -2 or x > 2. Interval notation: (-∞, -2) ∪ (2, +∞). Test point x=3: 9-4=5 > 0, confirmed. Test point x=0: 0-4=-4, not satisfied, confirmed.
Can a quadratic inequality have a solution of all real numbers?+
Yes. This occurs when the parabola stays entirely on the satisfying side of the x-axis. Example 1: x² + 1 > 0 - since D = -4 < 0 and a = 1 > 0, the parabola is always above the x-axis, so the solution is all real numbers. Example 2: -x² - 1 < 0 - same logic with a negative parabola always below the x-axis.
How do you verify the solution of a quadratic inequality?+
Pick a test point from inside your claimed solution interval and substitute it into the original inequality. It must make the inequality true. Also pick a test point outside the solution interval - it must make the inequality false. For x² - 5x + 6 < 0 with solution (2,3): test x=2.5: 6.25-12.5+6=-0.25 < 0, correct. Test x=0: 0-0+6=6 < 0, false, confirming x=0 is not in the solution.
What is the quadratic formula used to find boundary roots?+
The quadratic formula x = (-b ± √(b² - 4ac)) / (2a) solves ax² + bx + c = 0 for the boundary roots. The discriminant D = b² - 4ac determines root existence: D > 0 gives two real roots (two boundary points), D = 0 gives one repeated root (one boundary point), D < 0 gives no real roots (no boundary points, solution is all-or-nothing). The boundary roots divide the number line into regions that are then tested for the inequality.

How do you solve a quadratic inequality step by step?

Step 1: Rearrange to the form ax² + bx + c > 0 (or <, >=, <=). Step 2: Solve ax² + bx + c = 0 to find the boundary roots using the quadratic formula. Step 3: Note the sign of a and the direction of the inequality to identify which region satisfies it. Step 4: Write the solution set and interval notation.

What is interval notation for a quadratic inequality solution?

Interval notation uses parentheses () for excluded endpoints and brackets [] for included endpoints. For ax² + bx + c > 0 with a > 0 and roots x1 < x2, the solution is (-∞, x1) ∪ (x2, +∞). For < 0, the solution is (x1, x2). The ∪ symbol means 'union' (combine both parts).

What happens if the discriminant is negative in a quadratic inequality?

If D < 0, the quadratic has no real roots and the parabola never crosses the x-axis. If a > 0, the entire parabola is above the x-axis: ax² + bx + c > 0 for all real x (solution = all reals), and ax² + bx + c < 0 has no solution. If a < 0, it is reversed.

What does it mean when a quadratic inequality has no solution?

No solution means there are no real values of x that satisfy the inequality. For example, x² + 1 < 0 has no solution because x² + 1 is always positive (minimum value is 1 at x = 0). The solution set is the empty set, written as ∅ or {}.

How does the parabola direction affect the solution?

When a > 0 the parabola opens upward, so the expression is positive outside the roots and negative between them. When a < 0 the parabola opens downward, so the expression is positive between the roots and negative outside. This is why the same roots produce different solution sets for different signs of a.

What is the difference between strict and non-strict quadratic inequalities?

A strict inequality (> or <) excludes the boundary points where the expression equals zero. A non-strict inequality (>= or <=) includes them. In interval notation: strict uses parentheses at boundary points, non-strict uses brackets. Example: for roots 2 and 5, strict x² - 7x + 10 > 0 gives (-∞,2) ∪ (5,+∞) while non-strict >= 0 gives (-∞,2] ∪ [5,+∞).

How do you graph a quadratic inequality?

Graph y = ax² + bx + c as a parabola. For > 0 (or >= 0), shade the region above the x-axis (y > 0). For < 0 (or <= 0), shade the region below. The x-coordinates of the shaded region on the x-axis give the solution set. For strict inequalities, use open circles at the roots; for non-strict, use filled circles.

What is the solution when the quadratic has a repeated root?

With a repeated root x₀ (D = 0), the parabola just touches the x-axis at one point. If a > 0 and the inequality is > 0, the solution is all reals except x₀. If a > 0 and the inequality is >= 0, the solution is all real numbers. If a > 0 and < 0, there is no solution. If a > 0 and <= 0, the only solution is the single point x₀.

Can a quadratic inequality have a solution of all real numbers?

Yes. If a > 0 and D < 0 (no real roots, parabola always above x-axis), then ax² + bx + c > 0 is true for all real x. Similarly, if a < 0 and D < 0, then ax² + bx + c < 0 for all real x. These are the cases where the entire number line is the solution.

How do you write the solution set of x squared minus 5x plus 6 less than 0?

First solve x² - 5x + 6 = 0: x = 2 or x = 3. Since a = 1 > 0, the parabola opens upward, so x² - 5x + 6 < 0 between the roots. Solution set: 2 < x < 3. Interval notation: (2, 3). Test point x = 2.5: 6.25 - 12.5 + 6 = -0.25 < 0, confirmed.

What is the quadratic formula used in this calculator?

The quadratic formula x = (-b ± √(b² - 4ac)) / (2a) finds the roots of ax² + bx + c = 0. The discriminant D = b² - 4ac determines the nature of roots: D > 0 gives two distinct real roots, D = 0 gives one repeated root, D < 0 gives no real roots. The roots serve as the boundary points of the inequality solution.

How do you solve a quadratic inequality with no middle term?

For ax² + c > 0 (b = 0), the roots (if real) are x = ±√(-c/a). Example: x² - 9 > 0 has roots x = ±3 and since a = 1 > 0, the solution is x < -3 or x > 3, i.e. (-∞,-3) ∪ (3,+∞). Another example: x² + 4 > 0 has D = -16 < 0, so with a > 0 the expression is always positive: solution is all reals.