Quadratic Inequalities
Solve ax²+bx+c < 0 or > 0 by finding roots, then using a sign chart. Parabola opens up (a>0) or down (a<0). Solution is intervals where the quadratic is positive or negative. Used in profit, projectile, optimization problems.
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a>0: parabola opens up; positive outside roots, negative between. a<0: parabola opens down; negative outside roots, positive between. Include boundary (≤,≥) in solution; exclude for (<,>).
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Why: Quadratic inequalities model when profit is positive, when a projectile is above ground, or when a quantity exceeds a threshold. Sign chart shows where the parabola is above or below the x-axis.
How: Find roots of ax²+bx+c=0. If two roots: use sign chart on (−∞,r₁), (r₁,r₂), (r₂,∞). If one root: parabola touches axis. If no roots: quadratic is always positive or always negative (check a and vertex).
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📐 Examples — Click to Load
ax² + bx + c op 0
Critical Points & Vertex
Solution vs Excluded Intervals
📐 Calculation Steps
📊 Chart Interpretation
Bar chart: Displays vertex coordinates and roots. These critical points divide the number line into intervals for sign analysis.
Doughnut chart: Solution intervals (where the inequality holds) vs excluded intervals. Use for quick visual summary.
For educational and informational purposes only. Verify with a qualified professional.
🧮 Fascinating Math Facts
x²−4>0 → x<−2 or x>2. Parabola above axis outside roots.
— Example
Profit P(x)>0: solve quadratic inequality for revenue>cost.
— Business
📋 Key Takeaways
- • Parabola opens up/down: a > 0 → opens up; a < 0 → opens down
- • Sign analysis: Find roots, then test one value in each interval to determine sign
- • Connection to quadratic equation: Roots of ax²+bx+c=0 divide the number line into intervals
- • Real-world optimization: Projectile height, profit, revenue often modeled by quadratic inequalities
💡 Did You Know?
📖 How It Works
To solve x² - 5x + 6 ≥ 0: (1) Find roots: x = 2, x = 3. (2) Parabola opens up (a=1). (3) For ≥ 0, solution is x ≤ 2 or x ≥ 3.
Discriminant
Δ = b² - 4ac. Δ > 0: two roots; Δ = 0: one root; Δ < 0: no real roots. When Δ < 0, the quadratic is always positive or always negative.
Vertex Formula
Vertex at x = -b/(2a). Substitute to get y. The vertex is the max (a<0) or min (a>0) of the parabola.
Factoring Shortcut
When the quadratic factors nicely (e.g., x² - 5x + 6 = (x-2)(x-3)), the roots are immediate. Use the zero-product property.
📝 Worked Examples
Example 1: x² - 4 < 0
Factor: (x-2)(x+2) < 0. Roots at ±2. Parabola opens up, so negative between roots: -2 < x < 2. Interval: (-2, 2).
Example 2: x² - 5x + 6 ≥ 0
Factor: (x-2)(x-3) ≥ 0. Roots at 2 and 3. Opens up, so ≥ 0 outside: x ≤ 2 or x ≥ 3. Interval: (-∞, 2] ∪ [3, ∞).
Example 3: Projectile h(t) = -16t² + 48t > 0
Factor: -16t(t-3) > 0. Roots at 0 and 3. Parabola opens down, so positive between: 0 < t < 3. Ball is in the air for 3 seconds.
🌐 Real-World Optimization
Quadratic inequalities arise in physics (projectile motion), economics (profit/revenue), and engineering (stress, deflection). Setting P(x) ≥ 0 or h(t) > 0 finds valid ranges.
🔗 Related Concepts
Quadratic inequalities are solved by finding where the parabola crosses the x-axis (roots). The sign of the quadratic in each interval is determined by the parabola direction. This connects to solving quadratic equations and graphing parabolas.
📊 Table: Solution by Parabola Direction
| a > 0 (opens up) | a < 0 (opens down) |
|---|---|
| > 0 or ≥ 0: outside roots | > 0 or ≥ 0: between roots |
| < 0 or ≤ 0: between roots | < 0 or ≤ 0: outside roots |
🎯 Expert Tips
💡 Check a First
The sign of a determines parabola direction. a > 0: ∪ shape; a < 0: ∩ shape.
💡 No Real Roots
If Δ < 0, the quadratic never crosses zero. It is either always positive or always negative.
💡 Interval Notation
Use ( ) for strict < >; use [ ] for ≤ ≥. Union ∪ for "or" intervals.
💡 Real-World
Projectile height, profit, revenue: set inequality ≥ 0 or > 0 to find valid ranges.
❓ FAQ
When does the parabola open up vs down?
Opens up when a > 0, down when a < 0. The sign of a determines the shape.
How do I solve x² - 4 < 0?
Factor: (x-2)(x+2) < 0. Roots at ±2. Parabola opens up, so negative between roots: -2 < x < 2.
What is interval notation?
(a,b) means a < x < b; [a,b] means a ≤ x ≤ b. Use ∪ for union of intervals.
What is set builder notation?
{x | condition} means the set of x satisfying the condition. E.g., {x | 2 < x < 5}.
How does this connect to optimization?
Quadratic inequalities define feasible regions. The vertex often gives the optimal value.
What if the discriminant is negative?
No real roots. The quadratic is always positive (a>0) or always negative (a<0).
How do I solve (x-2)(x+3) ≥ 0?
Roots at -3 and 2. Parabola opens up (expand to get a=1). Solution: x ≤ -3 or x ≥ 2.
⚠️ Common Mistakes
1. Ignoring the sign of a: a > 0 means parabola opens up; a < 0 means it opens down. The solution intervals flip!
2. Wrong interval for ≥ or ≤: Include the roots when the inequality is ≥ or ≤. Use [ ] in interval notation.
3. Discriminant negative: No real roots means the quadratic never crosses zero. It is either always positive or always negative.
4. Double root (Δ=0): One repeated root. For > 0 or < 0, exclude that point; for ≥ or ≤, only that point may satisfy.
🔢 Quick Reference
📋 Step-by-Step Procedure
- Write the inequality in standard form: ax² + bx + c op 0. Ensure a ≠ 0.
- Find the roots by solving ax² + bx + c = 0. Use the quadratic formula or factoring.
- Determine parabola direction: a > 0 opens up, a < 0 opens down.
- Apply the rule: For a > 0, > 0 or ≥ 0 holds outside the roots; < 0 or ≤ 0 holds between. For a < 0, the intervals swap.
- Include or exclude roots based on ≤ vs < and ≥ vs >. Write the solution in interval notation.
The Bar chart displays vertex and roots. The Doughnut chart summarizes solution vs excluded regions.
💡 Remember: When a > 0, the parabola is ∪-shaped; when a < 0, it is ∩-shaped. The solution intervals depend on both the inequality direction and the parabola direction.
⚠️ Disclaimer: This calculator solves one-variable quadratic inequalities ax²+bx+c op 0. For two-variable inequalities y op ax²+bx+c, consider the parabola as a boundary. Educational use only.
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