A quadratic inequation (or inequality) is a mathematical statement that compares a quadratic polynomial to another value, typically zero, using an inequality symbol such as >, <, ≥, or ≤. Unlike a quadratic equation, which seeks specific values for the variable, an inequation seeks a range or set of values for which the statement is true.
Key Terms:
a ≠ 0. The coefficient 'a' determines the direction the parabola opens.b² - 4ac, which determines the number of real roots of the corresponding equation.ax² + bx + c = 0. These points divide the number line into intervals for testing the inequality.A quadratic inequation is visualized using a parabola defined by the function y = ax² + bx + c. The solution to an inequality like ax² + bx + c > 0 corresponds to the x-intervals where the parabola lies above the x-axis. The roots, x₁ and x₂, are the points where the parabola intersects the x-axis. The sign of the coefficient 'a' determines the parabola's orientation: if a > 0, it opens upwards (U-shape); if a < 0, it opens downwards (∩-shape).
| Property | Description |
|---|---|
| Interval Solutions | Solutions typically form continuous intervals or unions of intervals on the number line, representing a range of values. |
| Critical Points | The roots of the corresponding equation `ax² + bx + c = 0` act as critical points that divide the number line into regions where the expression has a consistent sign (+ or -). |
| Parabola Orientation | The sign of 'a' determines if the parabola opens up (a > 0) or down (a < 0), which dictates whether the solution for `> 0` is between or outside the roots. |
| Boundary Behavior | Strict inequalities (>, <) exclude the roots from the solution set (open intervals), while non-strict inequalities (≥, ≤) include them (closed intervals). |
The solution to a quadratic inequality is found by determining the intervals on the number line where the quadratic expression is positive or negative. This is achieved through the sign analysis method, which relies on the roots of the corresponding quadratic equation.
These roots, x₁ and x₂, are the critical points where the expression equals zero. They divide the number line into intervals, such as `(-∞, x₁)`, `(x₁, x₂)` and `(x₂, ∞)`.
Step 3: Substitute a test point from each interval into the expression `ax² + bx + c` to determine its sign (+ or -) for that entire interval.
If the inequality is `> 0` or `≥ 0`, choose the interval(s) where the test point yielded a positive result. If it is `< 0` or `≤ 0`, choose the interval(s) that yielded a negative result. Express the final answer using interval or set notation.
Quadratic inequalities are used to determine safety margins and design constraints. For example, finding the time intervals during which a projectile is above a certain height, or ensuring that the stress on a beam, modeled by a quadratic function, remains below a critical threshold.
They are applied to find profitable production ranges. A company's profit might be a quadratic function of production level; inequalities can determine the range of units to produce to ensure profit is above a certain minimum or to avoid losses.
In operations research, they help define feasible regions for resource allocation. Constraints in optimization problems can be quadratic, and inequalities delineate the set of valid solutions.
| Discriminant (Δ = b² - 4ac) | Number of Real Roots | Nature of Solution Set |
|---|---|---|
| `Δ > 0` | Two distinct real roots (x₁, x₂) | The solution is either the interval between the roots, `(x₁, x₂)` or the two intervals outside the roots, `(-∞, x₁) ∪ (x₂, ∞)`. |
| `Δ = 0` | One repeated real root (x₀) | The solution is typically all real numbers except the root, just the single root itself, or no solution, depending on the inequality. |
| `Δ < 0` | No real roots | The quadratic expression is always positive or always negative. The solution is either all real numbers (ℝ) or the empty set (∅). |
Forgetting to reverse the inequality sign when multiplying or dividing the entire inequation by a negative number. For instance, changing `-2x > 4` to `x > -2` is incorrect; it should be `x < -2`.
Incorrectly identifying the solution intervals. After finding the roots, one must test a point from each region. A common error is assuming the solution is always between the roots.
Confusing strict (>,