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.