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Last updated on September 29, 2025
A quadratic inequality compares a quadratic expression ax² + bx + c (where a ≠ 0) to zero using >, <, ≥, ≤ signs. They allow us to analyze intervals between roots and help determine if a given condition holds. Students should rewrite the inequality in standard form for solving, find the roots of ax² + bx + c = 0, and test sign intervals.
A quadratic inequality involves comparing a quadratic expression ax2 + bx + c, where a ≠ 0 to a number or another polynomial (degree ≤ 2), using >, <, ≥, or ≤. Unlike a quadratic equation (which equals something), inequalities yield ranges of solutions. Examples are:
Quadratic inequalities compare a quadratic expression to zero, using inequality symbols: greater than, less than, greater than or equal to, and less than or equal to. They are solved by finding roots and testing the signs of the expression over different intervals.
Type | General Form | Name |
Standard | ax2 + bx + c ≷ 0 | Positive/Negative/0m/0 |
No real roots | <0 | Always positive or negative |
Repeated roots | =0 | Touches the x-axis once |
Factored (simple/repeated) | (x - a)(x - b) . . .≷0 | Linear and repeated linear factors |
Rational | P(x)/Q(x)≷ 0 | Rational quadratic inequalities |
Compound | quad ≷ m quad | Two-sided/bounded inequality |
Step 1 - Rewrite the equation to express the inequality:
Example: x2 − 5x + 6 > 0.
Step 2 - Find roots by factoring or formula:
(x−2)(x−3)>0 → roots: 2, 3.
Step 3 - Create intervals:
(−∞,2), (2,3), (3,∞).
Step 4 - Test each interval by plugging in a sample value:
Step 5. Write a solution:
All x∈ (−∞,2) (3,∞).
The symbols > and < (greater than, less than) replace “=” in a quadratic equation to form a quadratic inequality. The format is
ax² + bx + c > 0 or ax² + bx + c < 0.
Examples:
Different methods used to solve quadratic inequalities are listed below:
Example:
X2 − 5x + 6 > 0 ⟹ (x − 2)(x − 3) > 0
(−∞,2), (2,3), (3,∞).
Testing the values in the expression, we get, x = 0: true; x = 2.5: false; x = 4: true.
Solution: x∈(−∞,2)∪(3,∞).
Example:
(x − 3)(x + 2) < 0.
x - 3 = 0, x = 3
x + 2 = 0, x = -2
So, the expression changes sign between x = -2 and x = 3
Now, set intervals using roots to divide the number line:
(-∞,-2), (-2,3), (3,∞)
Test each interval
x = -3: (x - 3) (x + 2) = (+)
x = 0: (x - 3) (x + 2) = (-)
x = 4: (x - 3) (x + 2) = (+)
The only region where the expression is negative is (-2, 3).
So, x∈(-2,3)
Solution: x∈(-2,3)
Example:
x2 + 5x + 6 ≥ 0.
The graph crosses at x = −2 and −3 and opens upward since the linear coefficient is positive.
So, the expression is ≥ 0 outside [−3,−2].
Solution: x∈(−∞, −3]∪[−2,∞).
The quadratic inequalities model helps us understand and solve real-world problems in the fields of physics, engineering, economics, and biology. Some of its applications are listed below:
Quadratic inequalities are challenging for some students, and common mistakes can create incorrect answers. Recognizing and avoiding these errors is important for perfect problem-solving.
For which values of x is x² − 4x + 3 > 0?
x∈ (-∞,1) ∪ (3,∞).
(x - 1)(x - 3) 3 > 0. Roots: x = 1, 3.
(-∞,1), (1,3), (3,∞).
x = 0: (0 - 1)(0 - 3) = (-1)(-3) 3 > 0, true
x = 2: (2 - 1)(2 - 3) = 1 · (-1) = -1 > 0, false
X = 4: (4 - 1)(4 - 3) = 3·1 = 3 > 0, true
Solve x² + 2x − 15 ≤ 0.
−5 ≤ x ≤ 3.
Factor to (x + 5)(x − 3).
The quadratic is non-positive between the roots.
Determine x such that 3x² − 12 ≤ 0.
−2 ≤ x ≤ 2
Rewrite as 3(x2 − 4) ≤ 0.
Hence, −2 ≤ x ≤ 2.
Find all x satisfying 2x²+ 7x + 6 < 0
−2 < x < −3/2
Factor (2x + 3)(x + 2) < 0.
The product is negative between the roots: −2 < x < −1.5.
Solve 4x² − 4x+ 1 ≥ 0.
All real x.
This is (2x − 1)2, which is always ≥ 0 for every x.