The Quadratic Formula

Introduction

The quadratic formula is a powerful tool that solves ANY quadratic equation, even when factoring is difficult or impossible.

For any quadratic equation in standard form: ax² + bx + c = 0

The solutions are given by:

x = (-b ± √(b² - 4ac)) / (2a)

This formula ALWAYS works for quadratic equations!

Understanding the Formula

Let's break down each part:

-b: The opposite of the coefficient of x

±: Plus-or-minus symbol (gives us two solutions)

√(b² - 4ac): The square root of the discriminant

2a: Twice the coefficient of x²

The discriminant: b² - 4ac This tells us about the nature of the solutions.

When to Use the Quadratic Formula

Use the quadratic formula when:

• The quadratic doesn't factor nicely
• You need exact decimal answers
• You want to be absolutely sure of the solutions
• Factoring seems too difficult

You might use other methods when:

• The equation factors easily
• It's in the form x² = k
• You're looking for quick integer solutions

Step-by-Step Process

Step 1: Write equation in standard form (ax² + bx + c = 0)

Step 2: Identify a, b, and c

Step 3: Substitute into the formula

Step 4: Simplify under the square root (the discriminant)

Step 5: Simplify the entire expression

Step 6: Write two solutions (one with +, one with -)

Step 7: Check your solutions

Example 1: Two Real Solutions

Solve: x² + 5x + 3 = 0

Step 1: Already in standard form

Step 2: Identify coefficients

• a = 1
• b = 5
• c = 3

Step 3: Substitute into formula x = (-5 ± √(5² - 4(1)(3))) / (2(1))

Step 4: Simplify discriminant x = (-5 ± √(25 - 12)) / 2 x = (-5 ± √13) / 2

Step 5: This is simplified (√13 cannot be simplified)

Step 6: Two solutions x = (-5 + √13) / 2 or x = (-5 - √13) / 2

Approximate values: x ≈ (-5 + 3.606) / 2 ≈ -0.697 x ≈ (-5 - 3.606) / 2 ≈ -4.303

Example 2: Simplifying Radicals

Solve: x² - 4x + 1 = 0

Coefficients: a = 1, b = -4, c = 1

Substitute: x = (4 ± √((-4)² - 4(1)(1))) / (2(1)) x = (4 ± √(16 - 4)) / 2 x = (4 ± √12) / 2

Simplify √12: √12 = √(4 · 3) = 2√3

x = (4 ± 2√3) / 2

Factor out 2: x = 2(2 ± √3) / 2 x = 2 ± √3

Solutions: x = 2 + √3 or x = 2 - √3

Example 3: When a ≠ 1

Solve: 2x² + 7x + 3 = 0

Coefficients: a = 2, b = 7, c = 3

Substitute: x = (-7 ± √(7² - 4(2)(3))) / (2(2)) x = (-7 ± √(49 - 24)) / 4 x = (-7 ± √25) / 4 x = (-7 ± 5) / 4

Two solutions: x = (-7 + 5) / 4 = -2/4 = -1/2 x = (-7 - 5) / 4 = -12/4 = -3

Check: This could have been factored as (2x + 1)(x + 3) = 0

Example 4: Rearranging First

Solve: 3x² = 5x + 2

Step 1: Standard form 3x² - 5x - 2 = 0

Coefficients: a = 3, b = -5, c = -2

Substitute: x = (5 ± √((-5)² - 4(3)(-2))) / (2(3)) x = (5 ± √(25 + 24)) / 6 x = (5 ± √49) / 6 x = (5 ± 7) / 6

Solutions: x = (5 + 7) / 6 = 12/6 = 2 x = (5 - 7) / 6 = -2/6 = -1/3

The Discriminant: b² - 4ac

The discriminant tells us the nature of solutions BEFORE we solve!

If b² - 4ac > 0: Two different real solutions

If b² - 4ac = 0: One repeated real solution (two equal solutions)

If b² - 4ac < 0: No real solutions (two complex solutions)

Using the Discriminant

Example 1: How many solutions does x² + 6x + 5 = 0 have?

Calculate discriminant: b² - 4ac = 6² - 4(1)(5) = 36 - 20 = 16

Since 16 > 0, there are two different real solutions.

We could solve: x = (-6 ± 4) / 2, giving x = -1 or x = -5

Example 2: How many solutions does x² - 4x + 4 = 0 have?

b² - 4ac = (-4)² - 4(1)(4) = 16 - 16 = 0

Since discriminant = 0, there is one repeated solution.

x = (4 ± 0) / 2 = 2

This is (x - 2)² = 0

Example 3: How many solutions does x² + 2x + 5 = 0 have?

b² - 4ac = 2² - 4(1)(5) = 4 - 20 = -16

Since -16 < 0, there are no real solutions.

The graph doesn't cross the x-axis.

One Repeated Solution

When the discriminant equals zero, you get one repeated root.

Example: x² - 10x + 25 = 0

a = 1, b = -10, c = 25

Discriminant: (-10)² - 4(1)(25) = 100 - 100 = 0

x = (10 ± 0) / 2 = 10/2 = 5

This is a perfect square: (x - 5)² = 0

Careful with Negative Signs

Be extra careful when b or c is negative!

Example: x² - 3x - 4 = 0

a = 1, b = -3, c = -4

x = (-(-3) ± √((-3)² - 4(1)(-4))) / (2(1)) x = (3 ± √(9 + 16)) / 2 x = (3 ± √25) / 2 x = (3 ± 5) / 2

x = 4 or x = -1

Common mistakes:

• Writing -b as -3 instead of -(-3) = 3
• Writing -4c as -4(-4) = 16, but forgetting the negative in front

Simplifying Square Roots

Always simplify radicals in your answer.

Example: If you get √50: √50 = √(25 · 2) = 5√2

Example: If you get √72: √72 = √(36 · 2) = 6√2

Example: x = (6 ± √48) / 4

Simplify √48 = √(16 · 3) = 4√3

x = (6 ± 4√3) / 4

Factor out 2: x = 2(3 ± 2√3) / 4 x = (3 ± 2√3) / 2

Applications: Projectile Motion

The height of a projectile is often modeled by: h = -16t² + v₀t + h₀

Example: A ball is thrown upward at 64 ft/s from height 6 ft. When does it hit the ground?

Equation: -16t² + 64t + 6 = 0

Divide by -2: 8t² - 32t - 3 = 0

Using quadratic formula: a = 8, b = -32, c = -3

t = (32 ± √(1024 + 96)) / 16 t = (32 ± √1120) / 16 t = (32 ± 4√70) / 16 t = (8 ± √70) / 4

t ≈ (8 + 8.37) / 4 ≈ 4.09 seconds (positive solution)

We reject the negative time solution.

Applications: Area Problems

Example: A rectangle has length 4 cm more than its width. Area is 60 cm². Find dimensions.

Let w = width Then w + 4 = length

w(w + 4) = 60 w² + 4w - 60 = 0

Using quadratic formula: w = (-4 ± √(16 + 240)) / 2 w = (-4 ± √256) / 2 w = (-4 ± 16) / 2

w = 6 or w = -10

Since width must be positive: w = 6 cm, length = 10 cm

Applications: Number Problems

Example: The sum of a number and its reciprocal is 13/6. Find the number.

Let x = the number Then 1/x = reciprocal

x + 1/x = 13/6

Multiply by 6x: 6x² + 6 = 13x Rearrange: 6x² - 13x + 6 = 0

x = (13 ± √(169 - 144)) / 12 x = (13 ± √25) / 12 x = (13 ± 5) / 12

x = 18/12 = 3/2 or x = 8/12 = 2/3

Both solutions work! (They're reciprocals of each other)

Checking Solutions

Substitute back into the original equation to verify.

Example: For x² - 5x + 6 = 0, we get x = 2 or x = 3

Check x = 2: 2² - 5(2) + 6 = 4 - 10 + 6 = 0 ✓ Check x = 3: 3² - 5(3) + 6 = 9 - 15 + 6 = 0 ✓

Common Mistakes to Avoid

1. Forgetting the negative sign in -b The formula is -b, not b!

2. Order of operations in discriminant Calculate b² first, then 4ac, then subtract

3. Not simplifying radicals √12 should be written as 2√3

4. Forgetting ± gives TWO solutions Don't just use + or just use -

5. Division errors Divide ENTIRE numerator by denominator

6. Sign errors with negative b or c Be extra careful: -(-3) = 3, -4(-2) = 8

Comparing Methods

Factoring:

• Fastest when it works
• Only works with nice integers
• x² + 5x + 6 = 0 → (x+2)(x+3) = 0

Quadratic Formula:

• Always works
• More calculation
• Exact answers with radicals
• x² + 5x + 3 = 0 → x = (-5 ± √13)/2

Completing the Square:

• Useful for deriving the formula
• Good for vertex form
• More steps than formula

Practice Strategy

Level 1: a = 1, perfect squares

• x² - 6x + 9 = 0

Level 2: a = 1, two solutions

• x² + 5x + 3 = 0

Level 3: a ≠ 1

• 2x² + 7x + 3 = 0

Level 4: Requires simplifying

• x² - 4x + 1 = 0

Level 5: Applications

• Projectile motion
• Area problems

Quick Reference

The Formula: x = (-b ± √(b² - 4ac)) / (2a)

The Discriminant: b² - 4ac

• Positive: 2 real solutions
• Zero: 1 solution
• Negative: 0 real solutions

Steps:

1. Standard form
2. Identify a, b, c
3. Substitute into formula
4. Simplify
5. Write both solutions
6. Check

Tips for Success

• Write the formula at the top of your work
• Show all substitution clearly
• Be careful with negative signs
• Simplify radicals completely
• Check discriminant first to know what to expect
• Always write two solutions (even if they're equal)
• Verify answers when possible
• Practice until the formula becomes automatic