MathIsimple
Lesson 4.2: Quadratic Formula & Discriminant

Master the Quadratic Formula & Discriminant

Discover the powerful quadratic formula that can solve any quadratic equation. Learn how the discriminant reveals the nature of solutions before you even calculate them.

Learning Objectives

Apply the quadratic formula to solve equations
Calculate and interpret the discriminant
Determine solution types from discriminant
Handle complex and irrational solutions

The Quadratic Formula

The Universal Formula

x = -b ± √(b² - 4ac)
2a

For any quadratic equation: ax² + bx + c = 0 (where a ≠ 0)

Formula Components

a: coefficient of x² (quadratic coefficient)

b: coefficient of x (linear coefficient)

c: constant term

±: indicates two possible solutions

√(b² - 4ac): the discriminant (determines solution type)

Step-by-Step Process

1

Identify coefficients

Find a, b, and c from ax² + bx + c = 0

2

Calculate discriminant

Find b² - 4ac to determine solution type

3

Substitute into formula

Replace a, b, c in the quadratic formula

4

Simplify and solve

Calculate both solutions using + and -

Example 1: Two Real Solutions

Solve: 2x² - 7x + 3 = 0

Step 1: Identify coefficients

a = 2, b = -7, c = 3

Step 2: Calculate discriminant

b² - 4ac = (-7)² - 4(2)(3) = 49 - 24 = 25

Since 25 > 0, there are two real solutions

Step 3: Apply quadratic formula

x = -(-7) ± √25
2(2)

x = 7 ± 5
4

Step 4: Calculate both solutions

x = (7 + 5)/4

x = 3

x = (7 - 5)/4

x = 1/2

Solutions: x = 3 or x = 1/2

Understanding the Discriminant

Discriminant = b² - 4ac

The discriminant tells us about the nature of solutions without solving the equation!

Discriminant > 0

Two real, distinct solutions

x = (-b + √D)/(2a)

x = (-b - √D)/(2a)

Discriminant = 0

One real solution (double root)

x = -b/(2a)

Discriminant < 0

No real solutions (complex)

x = (-b ± i√|D|)/(2a)

Example 2: One Real Solution

Solve: x² - 6x + 9 = 0

Step 1: Identify coefficients

a = 1, b = -6, c = 9

Step 2: Calculate discriminant

b² - 4ac = (-6)² - 4(1)(9) = 36 - 36 = 0

Since 0 = 0, there is one real solution

Step 3: Apply quadratic formula

x = -(-6) ± √0
2(1)

x = 6 ± 0
2

x = 6
2

Solution: x = 3 (double root)

Example 3: No Real Solutions

Solve: x² + 4x + 5 = 0

Step 1: Identify coefficients

a = 1, b = 4, c = 5

Step 2: Calculate discriminant

b² - 4ac = (4)² - 4(1)(5) = 16 - 20 = -4

Since -4 < 0, there are no real solutions

Step 3: Complex solutions (for reference)

x = -4 ± √(-4)
2(1)

x = -4 ± 2i
2

x = -2 ± i

No real solutions (complex: x = -2 ± i)

Common Mistakes to Avoid

❌ Mistake 1: Wrong sign for 'b'

In x² - 5x + 6 = 0, b = -5 (not 5). The formula uses -b, so -(-5) = +5

❌ Mistake 2: Forgetting to divide by 2a

Always divide the entire numerator by 2a, not just the discriminant part

❌ Mistake 3: Not simplifying the discriminant

Always simplify √(b² - 4ac) when possible (e.g., √25 = 5, not √25)

Practice Problems

Problem 1:

Solve: x² - 8x + 12 = 0

Show Solution

Discriminant: 64 - 48 = 16 > 0 (two real solutions)

x = (8 ± 4)/2

Solutions: x = 6 or x = 2

Problem 2:

Solve: 4x² - 4x + 1 = 0

Show Solution

Discriminant: 16 - 16 = 0 (one real solution)

x = 4/8 = 1/2

Solution: x = 1/2 (double root)

Problem 3:

Solve: x² + 2x + 3 = 0

Show Solution

Discriminant: 4 - 12 = -8 < 0 (no real solutions)

Complex solutions: x = -1 ± i√2

Advanced Applications

Optimization Problems

Maximum/Minimum Values

Use vertex formula: x = -b/(2a)

Example: Find maximum area of rectangle with perimeter 20

Complex Solutions

When Δ < 0

Solutions involve √(-1) = i

Example: x² + 4x + 5 = 0 has complex solutions

Real-world Applications

Engineering & Physics

Bridge Design

Finding optimal arch height

Calculating stress points and load distribution

Finance & Economics

Investment Analysis

Finding break-even points

Risk assessment and portfolio optimization

Technology & Computing

Algorithm Design

Time complexity analysis

Graphics programming and 3D modeling