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Understanding the Quadratic Formula
The quadratic formula is used to find the roots (solutions) of a quadratic equation of the form:
Where a, b, and c are coefficients, and a ≠ 0.
The Quadratic Formula
The solutions to the quadratic equation are given by:
Practical Example: Solving x² + 4x + 4 = 0
Here, a=1, b=4, c=4
x = [-4 ± √(16 - 16)] / 2
x = [-4 ± 0] / 2
x = -2
This equation has one real root (a repeated root) at x = -2
Components of the Quadratic Formula
- Discriminant (D): The expression under the square root (b² - 4ac) determines the nature of the roots
- Two solutions: The ± symbol means there are generally two solutions
- Real vs. complex: Roots can be real or complex numbers depending on the discriminant
Types of Solutions Based on Discriminant
| Discriminant | Nature of Roots | Example |
|---|---|---|
| D > 0 | Two distinct real roots | x² - 3x + 2 = 0 (roots at 1 and 2) |
| D = 0 | One real root (repeated) | x² - 2x + 1 = 0 (root at 1) |
| D < 0 | Two complex conjugate roots | x² + x + 1 = 0 (complex roots) |
Derivation of the Quadratic Formula
The quadratic formula can be derived by completing the square:
2. Divide by a: x² + (b/a)x + c/a = 0
3. Move constant term: x² + (b/a)x = -c/a
4. Complete the square: x² + (b/a)x + (b/2a)² = (b/2a)² - c/a
5. Factor left side: (x + b/2a)² = (b² - 4ac)/4a²
6. Take square root: x + b/2a = ±√(b² - 4ac)/2a
7. Solve for x: x = [-b ± √(b² - 4ac)] / 2a
Applications of Quadratic Equations
1. Physics: Projectile Motion
The height of a projectile over time follows a quadratic equation:
Where g is gravity, v₀ is initial velocity, h₀ is initial height.
Example: Ball Thrown Upward
A ball is thrown upward from ground level at 20 m/s:
h(t) = -4.9t² + 20t
To find when it hits the ground (h=0):
-4.9t² + 20t = 0
Solutions at t=0 (initial) and t≈4.08 seconds
2. Geometry: Area Problems
Many area and perimeter problems lead to quadratic equations.
Example: Rectangle Area
A rectangle has area 24 m² and length 2m more than width:
Let width = x, length = x+2
x(x+2) = 24 → x² + 2x - 24 = 0
Solutions: x=4 (valid) and x=-6 (invalid)
Dimensions: 4m × 6m
3. Economics: Profit Maximization
Profit functions are often quadratic, allowing calculation of maximum profit.
Example: Product Pricing
Profit P from selling x units at price p:
P = -2x² + 100x - 800
Maximum profit occurs at vertex: x = -b/2a = 25 units
Maximum profit = P(25) = $450
Common Mistakes to Avoid
- Forgetting the ± sign: The quadratic formula has two solutions
- Sign errors: Be careful with negative coefficients
- Dividing incorrectly: The denominator is 2a, not 2
- Order of operations: Calculate the discriminant (b² - 4ac) first
- Imaginary roots: Don't ignore complex solutions when they exist
Frequently Asked Questions
Q: What if a=0 in the quadratic equation?
A: If a=0, the equation becomes linear (bx + c = 0), not quadratic. The solution is simply x = -c/b.
Q: How do I handle complex roots?
A: Complex roots occur when the discriminant is negative. They come in conjugate pairs (a±bi). In many real-world applications, only real solutions are meaningful.
Q: What's the geometric interpretation of quadratic roots?
A: The roots represent the x-intercepts of the parabola y = ax² + bx + c. A positive discriminant means two intercepts, zero means one (vertex on x-axis), negative means none (parabola doesn't cross x-axis).
Q: Are there other methods to solve quadratic equations?
A: Yes, alternatives include factoring (when possible), completing the square, and graphing. However, the quadratic formula works for all cases.
Q: How is the quadratic formula related to the vertex form?
A: The vertex form y = a(x-h)² + k shows the vertex at (h,k). The x-coordinate of the vertex is h = -b/2a, which appears in the quadratic formula.