Lesson 4: Solving Quadratic Equations
Estimated time: 45-55 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Solve quadratic equations by factoring
- Apply the Zero Product Property to find solutions
- Use the quadratic formula to solve any quadratic equation
- Complete the square to solve quadratic equations
- Use the discriminant (b² - 4ac) to determine the number and type of solutions
- Choose the most efficient method for a given equation
- Solve application problems involving quadratic equations (projectile motion, area, etc.)
What Does It Mean to Solve a Quadratic Equation?
To solve a quadratic equation means to find all values of x that make the equation true. These values are called solutions, roots, or zeros of the equation.
Standard Form of a Quadratic Equation:
ax² + bx + c = 0
where a ≠ 0. Solutions are the x-values that make this equation true.
Graphical Interpretation
The solutions to ax² + bx + c = 0 are the x-intercepts of the parabola y = ax² + bx + c. These are the points where the graph crosses the x-axis (where y = 0).
- Two solutions: Parabola crosses x-axis twice
- One solution: Parabola touches x-axis at vertex
- No real solutions: Parabola doesn't touch x-axis
Method 1: Solving by Factoring
When a quadratic can be factored, this is often the quickest method. We use the Zero Product Property.
Zero Product Property: If AB = 0, then A = 0 or B = 0 (or both).
In other words, if the product of two factors equals zero, at least one of the factors must be zero.
Steps for Solving by Factoring
- Write the equation in standard form: ax² + bx + c = 0
- Factor the left side completely
- Set each factor equal to zero
- Solve each equation for x
- Check your solutions (optional but recommended)
Example 1: Basic Factoring
Solve: x² + 5x + 6 = 0
Solution:
Step 1: Already in standard form
Step 2: Factor
Find two numbers that multiply to 6 and add to 5: 2 and 3
(x + 2)(x + 3) = 0
Step 3: Set each factor = 0
x + 2 = 0 or x + 3 = 0
Step 4: Solve
x = -2 or x = -3
Answer: x = -2 or x = -3
Example 2: Difference of Squares
Solve: x² - 16 = 0
Solution:
This is a difference of squares: a² - b² = (a + b)(a - b)
x² - 16 = x² - 4²
(x + 4)(x - 4) = 0
Set each factor = 0:
x + 4 = 0 or x - 4 = 0
x = -4 or x = 4
Answer: x = -4 or x = 4
Example 3: Factoring with Leading Coefficient
Solve: 2x² + 7x + 3 = 0
Solution:
Factor by grouping or trial and error:
2x² + 7x + 3 = (2x + 1)(x + 3)
Set each factor = 0:
2x + 1 = 0 or x + 3 = 0
2x = -1 or x = -3
x = -1/2 or x = -3
Answer: x = -1/2 or x = -3
Example 4: Equation Not in Standard Form
Solve: x² = 3x + 10
Solution:
Step 1: Rearrange to standard form
x² - 3x - 10 = 0
Step 2: Factor
Find two numbers that multiply to -10 and add to -3: -5 and 2
(x - 5)(x + 2) = 0
Step 3: Solve
x - 5 = 0 or x + 2 = 0
x = 5 or x = -2
Answer: x = 5 or x = -2
Method 2: The Quadratic Formula
The quadratic formula works for ANY quadratic equation, even when factoring is difficult or impossible. This is the most powerful and general method.
The Quadratic Formula
For the equation ax² + bx + c = 0 (where a ≠ 0):
The ± symbol means there are typically two solutions:
x = (-b + √(b² - 4ac)) / (2a) and x = (-b - √(b² - 4ac)) / (2a)
Steps for Using the Quadratic Formula
- Write equation in standard form: ax² + bx + c = 0
- Identify a, b, and c
- Substitute into the formula
- Simplify carefully (especially under the square root)
- Simplify the final answer
Example 5: Using the Quadratic Formula
Solve: x² + 6x + 2 = 0
Solution:
Step 1 & 2: Identify a = 1, b = 6, c = 2
Step 3: Substitute into formula
x = (-6 ± √(6² - 4(1)(2))) / (2(1))
Step 4: Simplify
x = (-6 ± √(36 - 8)) / 2
x = (-6 ± √28) / 2
x = (-6 ± 2√7) / 2
Step 5: Simplify by factoring
x = 2(-3 ± √7) / 2
x = -3 ± √7
Answer: x = -3 + √7 or x = -3 - √7
(Approximately x ≈ -0.35 or x ≈ -5.65)
Example 6: Quadratic Formula with Leading Coefficient
Solve: 2x² - 5x - 3 = 0
Solution:
a = 2, b = -5, c = -3
Substitute:
x = (-(-5) ± √((-5)² - 4(2)(-3))) / (2(2))
x = (5 ± √(25 + 24)) / 4
x = (5 ± √49) / 4
x = (5 ± 7) / 4
Two solutions:
x = (5 + 7)/4 = 12/4 = 3
x = (5 - 7)/4 = -2/4 = -1/2
Answer: x = 3 or x = -1/2
Example 7: No Real Solutions
Solve: x² + 2x + 5 = 0
Solution:
a = 1, b = 2, c = 5
Substitute:
x = (-2 ± √(2² - 4(1)(5))) / (2(1))
x = (-2 ± √(4 - 20)) / 2
x = (-2 ± √(-16)) / 2
Problem: √(-16) is not a real number (can't take square root of a negative).
Answer: No real solutions
Graphically: The parabola y = x² + 2x + 5 doesn't cross the x-axis.
The Discriminant: Predicting the Number of Solutions
The expression b² - 4ac (the part under the square root in the quadratic formula) is called the discriminant. It tells us how many real solutions exist WITHOUT actually solving!
The Discriminant
(Delta, Δ, is the Greek letter commonly used for discriminant)
b² - 4ac > 0
Positive discriminant
Two distinct real solutions
Parabola crosses x-axis twice
Example: x² - 5x + 6 = 0
Δ = 25 - 24 = 1 > 0
b² - 4ac = 0
Zero discriminant
One real solution (repeated)
Parabola touches x-axis at vertex
Example: x² - 6x + 9 = 0
Δ = 36 - 36 = 0
b² - 4ac < 0
Negative discriminant
No real solutions
Parabola doesn't touch x-axis
Example: x² + 2x + 5 = 0
Δ = 4 - 20 = -16 < 0
Example 8: Using the Discriminant
Without solving, determine how many real solutions each equation has:
(a) x² + 4x + 3 = 0
a = 1, b = 4, c = 3
Δ = 4² - 4(1)(3) = 16 - 12 = 4
Since 4 > 0: Two real solutions
(b) 4x² - 12x + 9 = 0
a = 4, b = -12, c = 9
Δ = (-12)² - 4(4)(9) = 144 - 144 = 0
Since Δ = 0: One real solution
(c) 2x² + 3x + 4 = 0
a = 2, b = 3, c = 4
Δ = 3² - 4(2)(4) = 9 - 32 = -23
Since -23 < 0: No real solutions
Method 3: Completing the Square
Completing the square is a technique that rewrites a quadratic in vertex form, which reveals the solutions. While the quadratic formula is usually more efficient, completing the square is important for understanding and for deriving the quadratic formula itself.
Steps for Completing the Square
- If a ≠ 1, divide entire equation by a
- Move constant term to the right side
- Take half of the coefficient of x, square it, add to both sides
- Factor the left side as a perfect square
- Take square root of both sides (remember ±)
- Solve for x
Example 9: Completing the Square
Solve: x² + 8x + 7 = 0
Solution:
Step 1: a = 1 already, so skip this step
Step 2: Move constant to right
x² + 8x = -7
Step 3: Take half of 8, square it: (8/2)² = 16. Add to both sides
x² + 8x + 16 = -7 + 16
x² + 8x + 16 = 9
Step 4: Factor left side
(x + 4)² = 9
Step 5: Take square root of both sides
x + 4 = ±3
Step 6: Solve
x + 4 = 3 or x + 4 = -3
x = -1 or x = -7
Answer: x = -1 or x = -7
Example 10: Completing the Square with Leading Coefficient
Solve: 2x² - 12x + 10 = 0
Solution:
Step 1: Divide by 2
x² - 6x + 5 = 0
Step 2: Move constant
x² - 6x = -5
Step 3: Half of -6 is -3, square: (-3)² = 9
x² - 6x + 9 = -5 + 9
x² - 6x + 9 = 4
Step 4: Factor
(x - 3)² = 4
Step 5: Square root
x - 3 = ±2
Step 6: Solve
x - 3 = 2 or x - 3 = -2
x = 5 or x = 1
Answer: x = 5 or x = 1
Which Method Should You Use?
Method Selection Guide
Use Factoring When:
- Equation factors easily
- Numbers are small integers
- You can spot patterns (difference of squares, perfect square trinomials)
Fastest method when applicable!
Use Quadratic Formula When:
- Equation doesn't factor easily
- Coefficients are large or messy
- You need exact answers with radicals
- You're unsure (it always works!)
Most versatile method
Use Completing the Square When:
- Converting to vertex form
- Deriving the quadratic formula
- Specific instructions require it
Less common for solving
Use Discriminant When:
- You only need to know HOW MANY solutions
- Determining if solutions are real
- Don't need actual solution values
Quick analysis tool
Applications of Quadratic Equations
Quadratic equations appear in many real-world situations: projectile motion, area problems, profit maximization, and more.
Example 11: Projectile Motion
A ball is thrown upward with an initial velocity of 48 feet per second from a height of 6 feet. Its height h (in feet) after t seconds is given by:
h = -16t² + 48t + 6
(a) When will the ball hit the ground?
(b) When will the ball reach its maximum height?
Solution (a): The ball hits the ground when h = 0
0 = -16t² + 48t + 6
Using quadratic formula: a = -16, b = 48, c = 6
t = (-48 ± √(48² - 4(-16)(6))) / (2(-16))
t = (-48 ± √(2304 + 384)) / (-32)
t = (-48 ± √2688) / (-32)
t = (-48 ± 51.85) / (-32)
Two values:
t = (-48 + 51.85)/(-32) ≈ -0.12 (reject: negative time)
t = (-48 - 51.85)/(-32) ≈ 3.12 seconds
Answer (a): The ball hits the ground after about 3.12 seconds
Solution (b): Maximum occurs at vertex
t = -b/(2a) = -48/(2(-16)) = -48/(-32) = 1.5 seconds
Maximum height:
h = -16(1.5)² + 48(1.5) + 6
h = -36 + 72 + 6 = 42 feet
Answer (b): Maximum height of 42 feet at t = 1.5 seconds
Example 12: Area Problem
A rectangular garden has length 4 feet more than its width. If the area is 60 square feet, find the dimensions.
Solution:
Let w = width in feet
Then length = w + 4
Area equation:
w(w + 4) = 60
w² + 4w = 60
w² + 4w - 60 = 0
Factor:
(w + 10)(w - 6) = 0
w = -10 or w = 6
Reject w = -10 (width can't be negative)
Answer:
Width = 6 feet
Length = 6 + 4 = 10 feet
Check: 6 × 10 = 60
Example 13: Consecutive Integers
Find two consecutive positive integers whose product is 182.
Solution:
Let n = first integer
Then n + 1 = next consecutive integer
Equation:
n(n + 1) = 182
n² + n = 182
n² + n - 182 = 0
Factor:
(n + 14)(n - 13) = 0
n = -14 or n = 13
Since we need positive integers, n = 13
Answer: The integers are 13 and 14
Check: 13 × 14 = 182
Check Your Understanding
Try these questions to see if you've grasped the key concepts:
1. Solve by factoring: x² - 9x + 20 = 0
Factor: (x - 4)(x - 5) = 0
x - 4 = 0 or x - 5 = 0
x = 4 or x = 5
2. Solve using the quadratic formula: x² + 4x - 1 = 0
a = 1, b = 4, c = -1
x = (-4 ± √(16 - 4(1)(-1))) / 2
x = (-4 ± √20) / 2
x = (-4 ± 2√5) / 2
x = -2 ± √5
x = -2 + √5 or x = -2 - √5
(Approximately x ≈ 0.24 or x ≈ -4.24)
3. Use the discriminant to determine the number of real solutions: 3x² - 2x + 5 = 0
a = 3, b = -2, c = 5
Δ = b² - 4ac
Δ = (-2)² - 4(3)(5)
Δ = 4 - 60
Δ = -56
Since Δ < 0: No real solutions
4. Solve: 2x² - 8x = 0 (Hint: Factor out common term first)
Factor out 2x:
2x(x - 4) = 0
2x = 0 or x - 4 = 0
x = 0 or x = 4
5. Solve by completing the square: x² + 10x + 9 = 0
x² + 10x = -9
Half of 10 is 5, square: 25
x² + 10x + 25 = -9 + 25
(x + 5)² = 16
x + 5 = ±4
x = -5 + 4 = -1 or x = -5 - 4 = -9
x = -1 or x = -9
6. A rectangle's length is 3 cm more than twice its width. If the area is 27 cm², find the width.
Let w = width
Length = 2w + 3
Area: w(2w + 3) = 27
2w² + 3w = 27
2w² + 3w - 27 = 0
Factor: (2w + 9)(w - 3) = 0
w = -9/2 (reject) or w = 3
Width = 3 cm
Length = 2(3) + 3 = 9 cm
Key Takeaways
- Three main methods: Factoring, Quadratic Formula, Completing the Square
- Zero Product Property: If AB = 0, then A = 0 or B = 0
- Quadratic Formula: x = (-b ± √(b² - 4ac)) / (2a) works for ALL quadratics
- Discriminant (Δ = b² - 4ac): Determines number of real solutions
- Δ > 0: Two real solutions
- Δ = 0: One real solution
- Δ < 0: No real solutions
- Factoring: Fastest when equation factors easily
- Completing the square: Useful for converting to vertex form
- Always write equation in standard form (ax² + bx + c = 0) first
- In applications, reject negative or unrealistic solutions
Module Complete!
Congratulations!
You've completed Module 2: Linear & Quadratic Functions! You've learned about linear equations, modeling, quadratic functions, and solving quadratics.
Next, test your knowledge with the module practice problems and quiz.
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