Quadratic Equations and Inequalities
Complete Notes & Formulae for Eleventh Grade (Algebra 2)
1. Solve Quadratic Equations Using Square Roots
When to Use:
Use this method when the equation has NO linear term (no x term)
General form: \( ax^2 + c = 0 \) or \( a(x - h)^2 + k = 0 \)
Square Root Property:
\[ \text{If } x^2 = k, \text{ then } x = \pm\sqrt{k} \]
Steps:
1. Isolate the squared term on one side
2. Take the square root of both sides
3. Remember to use ± (plus or minus)
4. Simplify
Examples:
Example 1: \( x^2 = 16 \)
\( x = \pm\sqrt{16} = \pm 4 \)
Solutions: \( x = 4 \) or \( x = -4 \)
Example 2: \( 3x^2 - 27 = 0 \)
Step 1: \( 3x^2 = 27 \)
Step 2: \( x^2 = 9 \)
Step 3: \( x = \pm 3 \)
Solutions: \( x = 3 \) or \( x = -3 \)
Example 3: \( (x - 5)^2 = 9 \)
\( x - 5 = \pm 3 \)
\( x - 5 = 3 \) or \( x - 5 = -3 \)
Solutions: \( x = 8 \) or \( x = 2 \)
2. Solve Using Zero Product Property and Factoring
Zero Product Property:
\[ \text{If } ab = 0, \text{ then } a = 0 \text{ or } b = 0 \]
If the product of two factors equals zero, then at least one of the factors must equal zero
Solving by Factoring:
Steps:
1. Write equation in standard form: \( ax^2 + bx + c = 0 \)
2. Factor the quadratic expression
3. Set each factor equal to zero
4. Solve each equation
Examples:
Example 1: \( x^2 + 5x + 6 = 0 \)
Factor: \( (x + 2)(x + 3) = 0 \)
Set each factor to zero: \( x + 2 = 0 \) or \( x + 3 = 0 \)
Solutions: \( x = -2 \) or \( x = -3 \)
Example 2: \( 2x^2 - 5x - 3 = 0 \)
Factor: \( (2x + 1)(x - 3) = 0 \)
\( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \)
\( x - 3 = 0 \) → \( x = 3 \)
Solutions: \( x = -\frac{1}{2} \) or \( x = 3 \)
3. Complete the Square
Completing the Square Formula:
To complete the square for \( x^2 + bx \), add \( \left(\frac{b}{2}\right)^2 \)
\[ x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2 \]
Steps to Complete the Square:
For \( ax^2 + bx + c = 0 \):
1. If \( a \neq 1 \), divide entire equation by \( a \)
2. Move constant term to right side
3. Take half of the coefficient of x, square it, add to both sides
4. Write left side as a perfect square
5. Solve using square root property
Example:
Solve: \( x^2 + 6x + 2 = 0 \)
Step 1: Move constant: \( x^2 + 6x = -2 \)
Step 2: Half of 6 is 3, square it: \( 3^2 = 9 \)
Step 3: Add 9 to both sides: \( x^2 + 6x + 9 = -2 + 9 \)
Step 4: \( (x + 3)^2 = 7 \)
Step 5: \( x + 3 = \pm\sqrt{7} \)
Step 6: \( x = -3 \pm \sqrt{7} \)
Solutions: \( x = -3 + \sqrt{7} \) or \( x = -3 - \sqrt{7} \)
4. Solve Using the Quadratic Formula
The Quadratic Formula:
For any quadratic equation \( ax^2 + bx + c = 0 \) where \( a \neq 0 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Works for ALL quadratic equations!
Steps:
1. Write equation in standard form: \( ax^2 + bx + c = 0 \)
2. Identify values of \( a, b, \) and \( c \)
3. Substitute into formula (use parentheses!)
4. Simplify under the radical first
5. Simplify completely
Example:
Solve: \( 2x^2 + 3x - 5 = 0 \)
Identify: \( a = 2, b = 3, c = -5 \)
Substitute: \( x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)} \)
Simplify: \( x = \frac{-3 \pm \sqrt{9 + 40}}{4} \)
\( x = \frac{-3 \pm \sqrt{49}}{4} \)
\( x = \frac{-3 \pm 7}{4} \)
\( x = \frac{-3 + 7}{4} = 1 \) or \( x = \frac{-3 - 7}{4} = -\frac{5}{2} \)
Solutions: \( x = 1 \) or \( x = -\frac{5}{2} \)
5. Using the Discriminant
The Discriminant:
The expression under the square root in the quadratic formula
\[ D = b^2 - 4ac \]
The discriminant tells us the number and type of solutions WITHOUT actually solving!
Discriminant Rules:
| Discriminant Value | Number of Solutions | Type of Solutions |
|---|---|---|
| \( D > 0 \) | Two solutions | Two different real solutions |
| \( D = 0 \) | One solution | One real solution (repeated root) |
| \( D < 0 \) | No real solutions | Two complex solutions |
Examples:
Equation: \( x^2 + 5x + 6 = 0 \)
\( D = 5^2 - 4(1)(6) = 25 - 24 = 1 > 0 \)
Two different real solutions
Equation: \( x^2 - 6x + 9 = 0 \)
\( D = (-6)^2 - 4(1)(9) = 36 - 36 = 0 \)
One real solution (repeated root)
Equation: \( x^2 + 2x + 5 = 0 \)
\( D = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0 \)
No real solutions (two complex solutions)
6. Solve Quadratic Equations: Word Problems
Steps for Word Problems:
1. Read the problem carefully and identify what you need to find
2. Define variables
3. Write an equation based on the problem
4. Solve the equation using appropriate method
5. Check if solution makes sense in context
6. Answer the question with proper units
Common Types:
• Area Problems
Rectangle: \( A = lw \), Square: \( A = s^2 \)
• Projectile Motion
\( h(t) = -16t^2 + v_0t + h_0 \) (feet) or \( h(t) = -4.9t^2 + v_0t + h_0 \) (meters)
• Number Problems
Consecutive integers, products, sums
• Work and Rate Problems
Time, rate, distance relationships
Example:
Problem: A ball is thrown upward with initial velocity 48 ft/s from a height of 6 feet. When does it hit the ground?
Equation: \( h(t) = -16t^2 + 48t + 6 \)
Ground means \( h = 0 \): \( -16t^2 + 48t + 6 = 0 \)
Using quadratic formula: \( t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(6)}}{2(-16)} \)
\( t = \frac{-48 \pm \sqrt{2304 + 384}}{-32} = \frac{-48 \pm \sqrt{2688}}{-32} \)
\( t \approx 3.12 \) seconds (positive solution)
Answer: The ball hits the ground after approximately 3.12 seconds
7. Solve Quadratic Inequalities
Types of Quadratic Inequalities:
• \( ax^2 + bx + c > 0 \)
• \( ax^2 + bx + c < 0 \)
• \( ax^2 + bx + c \geq 0 \)
• \( ax^2 + bx + c \leq 0 \)
Method 1: Algebraic (Test Point Method)
Steps:
1. Solve the related equation \( ax^2 + bx + c = 0 \) to find boundary points
2. Plot boundary points on number line (open circle for <, >, closed for ≤, ≥)
3. Test points divide line into regions
4. Test a value from each region in original inequality
5. Shade regions where inequality is TRUE
Example:
Solve: \( x^2 - 5x + 6 < 0 \)
Step 1: Solve \( x^2 - 5x + 6 = 0 \)
Factor: \( (x - 2)(x - 3) = 0 \)
Boundary points: \( x = 2 \) and \( x = 3 \)
Step 2: Test regions:
Region 1 (\( x < 2 \)): Test \( x = 0 \) → \( 6 < 0 \) FALSE
Region 2 (\( 2 < x < 3 \)): Test \( x = 2.5 \) → \( -0.75 < 0 \) TRUE ✓
Region 3 (\( x > 3 \)): Test \( x = 4 \) → \( 2 < 0 \) FALSE
Solution: \( 2 < x < 3 \) or \( (2, 3) \)
8. Graph Solutions to Quadratic Inequalities
Method 2: Graphical Method
Steps:
1. Graph the related quadratic function \( y = ax^2 + bx + c \)
2. Use solid parabola for ≤ or ≥, dashed for < or >
3. Shade the appropriate region:
• \( y > ax^2 + bx + c \): shade ABOVE parabola
• \( y < ax^2 + bx + c \): shade BELOW parabola
• \( y \geq ax^2 + bx + c \): shade ABOVE and ON parabola
• \( y \leq ax^2 + bx + c \): shade BELOW and ON parabola
Key Points:
• Find vertex for accurate graphing
• Find x-intercepts (zeros) if they exist
• Determine if parabola opens up (\( a > 0 \)) or down (\( a < 0 \))
• Test a point to verify shading direction
System of Inequalities:
When graphing a system:
• Graph each inequality separately
• Solution is where ALL shaded regions OVERLAP
9. Choosing the Best Method
| Method | When to Use |
|---|---|
| Square Roots | No x term, easy to isolate \( x^2 \) |
| Factoring | Expression is easily factorable |
| Completing Square | Converting to vertex form, \( a = 1 \) |
| Quadratic Formula | Works for ALL quadratics, non-factorable |
10. Quick Reference Summary
Key Formulas:
Standard Form: \( ax^2 + bx + c = 0 \)
Quadratic Formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Discriminant: \( D = b^2 - 4ac \)
\( D > 0 \): Two real solutions
\( D = 0 \): One real solution
\( D < 0 \): No real solutions
Zero Product Property:
If \( ab = 0 \), then \( a = 0 \) or \( b = 0 \)
Square Root Property:
If \( x^2 = k \), then \( x = \pm\sqrt{k} \)
Completing the Square:
Add \( \left(\frac{b}{2}\right)^2 \) to \( x^2 + bx \)
📚 Study Tips
✓ Always write quadratics in standard form first
✓ Check if equation can be factored before using quadratic formula
✓ Use discriminant to determine number of solutions quickly
✓ Remember ± when taking square roots
✓ For inequalities, test points to verify your solution