Quadratic Equations: Complete Guide
What is a Quadratic Equation?
A quadratic equation is a second-degree polynomial equation in a single variable, typically written in the form:
where:
- a is the coefficient of x² (a ≠ 0)
- b is the coefficient of x
- c is the constant term
The solutions to a quadratic equation are called its roots or zeros.
The Discriminant
The discriminant (Δ) of a quadratic equation determines the nature of its roots:
| Discriminant | Nature of Roots | Example |
|---|---|---|
| Δ > 0 | Two distinct real roots | x² - 5x + 6 = 0 |
| Δ = 0 | One repeated real root (equal roots) | x² - 6x + 9 = 0 |
| Δ < 0 | Two complex conjugate roots | x² + x + 1 = 0 |
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is:
Examples of Quadratic Equations
1. 2x² - 7x + 3 = 0
Here, a = 2, b = -7, c = 3
2. x² + 5x = 0
Here, a = 1, b = 5, c = 0
3. 4x² = 9
Rearranging to standard form: 4x² - 9 = 0
Here, a = 4, b = 0, c = -9
Note: Not all quadratic equations initially appear in standard form. Always arrange terms so that one side equals zero and the coefficient of x² is non-zero.
Alternative Forms
Factored Form
where r₁ and r₂ are the roots of the equation.
Vertex Form
where (h, k) is the vertex of the parabola.
Converting Between Forms
Standard to Factored:
x² - 5x + 6 = 0
Factoring: (x - 2)(x - 3) = 0
Standard to Vertex:
x² - 6x + 8 = 0
Complete the square: (x - 3)² - 9 + 8 = 0
Simplify: (x - 3)² - 1 = 0
Vertex form with h = 3, k = 1
Methods for Solving Quadratic Equations
There are several methods to solve quadratic equations:
1. Factoring Method
If you can factor the quadratic expression, this method is straightforward:
- Arrange the equation in standard form: ax² + bx + c = 0
- Factor the expression into (px + q)(rx + s) = 0
- Set each factor equal to zero and solve for x
Example: Solve x² - 7x + 12 = 0
Factor: (x - 3)(x - 4) = 0
Set each factor to zero:
x - 3 = 0 → x = 3
x - 4 = 0 → x = 4
2. Quadratic Formula
Always works for any quadratic equation:
Example: Solve 2x² - 5x - 3 = 0
Here: a = 2, b = -5, c = -3
x = (5 ± √(25 + 24)) / 4
x = (5 ± √49) / 4
x = (5 ± 7) / 4
3. Completing the Square
Transforms the quadratic into vertex form:
- Arrange the equation as ax² + bx + c = 0
- Divide all terms by a (if a ≠ 1)
- Move the constant term to the right side
- Add (b/2)² to both sides
- Rewrite the left side as a perfect square
- Solve for x
Example: Solve x² - 6x + 8 = 0
Rearrange: x² - 6x = -8
Add (b/2)² = (-6/2)² = 9 to both sides: x² - 6x + 9 = -8 + 9
Rewrite as perfect square: (x - 3)² = 1
Take square root of both sides: x - 3 = ±1
4. Square Root Method
Used when there's no linear term (b = 0):
Example: Solve 3x² - 12 = 0
Rearrange: 3x² = 12
Divide by 3: x² = 4
Take square root of both sides: x = ±2
5. Graphical Method
Find the x-intercepts of the parabola y = ax² + bx + c:
Example: Solve x² - 2x - 8 = 0 graphically
Graph y = x² - 2x - 8 and find where y = 0
Special Cases
Pure Quadratics (b = 0)
When the equation is in the form ax² + c = 0:
Example: Solve 4x² - 25 = 0
Rearrange: 4x² = 25
Divide by 4: x² = 6.25
Take square root: x = ±2.5
When c = 0
The equation becomes ax² + bx = 0, which can be factored as x(ax + b) = 0
Example: Solve 3x² + 15x = 0
Factor out x: x(3x + 15) = 0
Set each factor to zero:
x = 0 or 3x + 15 = 0
Real-World Applications
1. Projectile Motion
The height of an object thrown upward with initial velocity v₀ from height h₀ is:
Example: Finding Time of Impact
A ball is thrown upward at 20 m/s from a 15m tall building. When will it hit the ground?
h(t) = -4.9t² + 20t + 15
When the ball hits the ground, h(t) = 0:
-4.9t² + 20t + 15 = 0
2. Area Problems
Example: Fencing a Garden
A farmer has 100m of fencing to enclose a rectangular garden. What dimensions maximize the area?
Let w = width and l = length. Then:
Perimeter: 2w + 2l = 100
Solving for l: l = (100 - 2w)/2 = 50 - w
Area: A = w × l = w(50 - w) = 50w - w²
To maximize area, complete the square:
A = -w² + 50w = -(w² - 50w) = -(w² - 50w + 625 - 625) = -(w - 25)² + 625
3. Economics: Supply and Demand
Example: Finding Equilibrium
Suppose supply and demand are modeled by:
Supply: p = 2q² + 5
Demand: p = 100 - q²
At equilibrium, supply equals demand:
2q² + 5 = 100 - q²
3q² = 95
q² = 31.67
Algebraic Applications
Number Problems
Example: Finding Two Numbers
Find two numbers whose sum is 12 and product is 35.
Let the numbers be x and y.
x + y = 12
xy = 35
From the first equation: y = 12 - x
Substitute into the second: x(12 - x) = 35
12x - x² = 35
x² - 12x + 35 = 0
Geometry Applications
Example: Dimensions of a Rectangle
The length of a rectangle is 5 cm more than its width. If the area is 84 cm², find the dimensions.
Let w = width and l = length.
l = w + 5
Area: w × l = 84
w(w + 5) = 84
w² + 5w = 84
w² + 5w - 84 = 0
