Lesson 1 • 45 min read
Quadratic Equations
In This Lesson
1Standard Form
ax² + bx + c = 0
where a ≠ 0
a
Leading coefficient
(coefficient of x²)
b
Linear coefficient
(coefficient of x)
c
Constant term
(no variable)
Examples:
x² - 5x + 6 = 0
a=1, b=-5, c=6
2x² + 3x - 1 = 0
a=2, b=3, c=-1
2Solving by Factoring
Zero Product Property
If ab = 0, then a = 0 or b = 0
Steps
- Write equation in standard form (= 0)
- Factor the quadratic expression
- Set each factor equal to zero
- Solve each equation
Example 1: x² - 5x + 6 = 0
(x - 2)(x - 3) = 0 ← Factor
x - 2 = 0 or x - 3 = 0 ← Set each = 0
x = 2 or x = 3
Example 2: x² - 9 = 0 (Difference of Squares)
(x + 3)(x - 3) = 0
x = -3 or x = 3
Example 3: 2x² + 5x - 3 = 0
(2x - 1)(x + 3) = 0 ← Factor with a ≠ 1
2x - 1 = 0 or x + 3 = 0
x = 1/2 or x = -3
3Square Root Property
If x² = k, then x = ±√k
Use when there's no linear term (no bx)
Example 1: x² = 16
x = ±√16
x = ±4 (x = 4 or x = -4)
Example 2: x² - 25 = 0
x² = 25 ← Add 25
x = ±5
Example 3: (x - 3)² = 49
x - 3 = ±7 ← Square root both sides
x = 3 + 7 = 10 or x = 3 - 7 = -4
x = 10 or x = -4
⚠️ Don't Forget the ±
Both positive and negative roots are solutions! √16 = 4, but x² = 16 has two solutions: 4 and -4.
4Completing the Square
Steps
- Move constant to right side: x² + bx = -c
- Take half of b, square it: (b/2)²
- Add this value to both sides
- Factor left side as a perfect square
- Take square root of both sides (don't forget ±)
- Solve for x
Example: x² + 6x - 7 = 0
Step 1: x² + 6x = 7
Step 2: (6/2)² = 9
Step 3: x² + 6x + 9 = 7 + 9
Step 4: (x + 3)² = 16
Step 5: x + 3 = ±4
Step 6: x = -3 + 4 = 1 or x = -3 - 4 = -7
x = 1 or x = -7
Example 2: x² - 4x - 5 = 0
x² - 4x = 5
(-4/2)² = 4 ← Half of -4, squared
x² - 4x + 4 = 5 + 4
(x - 2)² = 9
x - 2 = ±3
x = 5 or x = -1
5Quadratic Formula
x = (-b ± √(b² - 4ac)) / 2a
Works for ALL quadratic equations!
The Discriminant: b² - 4ac
b² - 4ac > 0
Two real solutions
b² - 4ac = 0
One real solution
b² - 4ac < 0
No real solutions
Example: 2x² - 5x - 3 = 0
a = 2, b = -5, c = -3
x = (-(-5) ± √((-5)² - 4(2)(-3))) / 2(2)
x = (5 ± √(25 + 24)) / 4
x = (5 ± √49) / 4
x = (5 ± 7) / 4
x = (5 + 7)/4 = 3 or x = (5 - 7)/4 = -1/2
x = 3 or x = -1/2
Example with No Real Solution: x² + x + 1 = 0
a = 1, b = 1, c = 1
Discriminant = 1² - 4(1)(1) = 1 - 4 = -3
Since -3 < 0, no real solutions
Which Method to Use?
Factoring: When the equation factors easily
Square Root: When there's no bx term
Completing Square: When you need vertex form
Quadratic Formula: Always works, especially for complex equations