Quadratic Equations: The Complete Guide to Solving and Understanding
Quadratic equations are among the most important and frequently encountered equations in mathematics. From calculating projectile trajectories to optimizing business profits, quadratic equations appear throughout science, engineering, and everyday problem-solving. This comprehensive guide will teach you everything you need to know about understanding, solving, and applying quadratic equations.
What is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2.
Standard Form
The standard form of a quadratic equation is:
ax² + bx + c = 0
Where:
- a, b, c are constants (real numbers)
- a ≠ 0 (if a = 0, it becomes a linear equation)
- x is the variable (unknown)
Examples
- x² - 5x + 6 = 0 (a=1, b=-5, c=6)
- 2x² + 3x - 2 = 0 (a=2, b=3, c=-2)
- x² - 16 = 0 (a=1, b=0, c=-16)
- 3x² + 12x = 0 (a=3, b=12, c=0)
Other Forms
Vertex Form: y = a(x - h)² + k
- (h, k) is the vertex of the parabola
Factored Form: y = a(x - r₁)(x - r₂)
- r₁ and r₂ are the roots (solutions)
Understanding Solutions (Roots)
The solutions to a quadratic equation are called roots, zeros, or x-intercepts. A quadratic equation can have:
- Two distinct real roots (discriminant > 0)
- One repeated real root (discriminant = 0)
- Two complex roots (discriminant < 0)
The discriminant is: Δ = b² - 4ac
Method 1: Solving by Factoring
Factoring is often the fastest method when the quadratic can be easily factored.
The Zero Product Property
If AB = 0, then either A = 0 or B = 0 (or both).
This means: if (x - r₁)(x - r₂) = 0, then x = r₁ or x = r₂
Simple Factoring (a = 1)
Example 1: Solve x² + 5x + 6 = 0
Step 1: Factor the left side
Find two numbers that multiply to 6 and add to 5: 2 and 3
(x + 2)(x + 3) = 0
Step 2: Apply zero product property
x + 2 = 0 OR x + 3 = 0
x = -2 OR x = -3
Solutions: x = -2, x = -3
Example 2: Solve x² - 7x + 12 = 0
Step 1: Factor
Numbers: -3 and -4 (multiply to 12, add to -7)
(x - 3)(x - 4) = 0
Step 2: Solve
x = 3 or x = 4
Solutions: x = 3, x = 4
Factoring with a ≠ 1
Example 3: Solve 2x² + 7x + 3 = 0
Step 1: Factor using AC method
ac = 2 × 3 = 6
Find numbers that multiply to 6 and add to 7: 1 and 6
Step 2: Rewrite and group
2x² + x + 6x + 3 = 0
x(2x + 1) + 3(2x + 1) = 0
(2x + 1)(x + 3) = 0
Step 3: Solve
2x + 1 = 0 OR x + 3 = 0
x = -1/2 OR x = -3
Solutions: x = -1/2, x = -3
Special Cases
Difference of Squares:
x² - 9 = 0 (x + 3)(x - 3) = 0 x = 3 or x = -3
Perfect Square:
x² + 6x + 9 = 0 (x + 3)² = 0 x = -3 (double root)
Use our quadratic factorization calculator to factor and solve quadratic equations instantly.
Method 2: The Quadratic Formula
The quadratic formula works for any quadratic equation, even when factoring is difficult or impossible.
The Formula
For ax² + bx + c = 0:
x = [-b ± √(b² - 4ac)] / (2a)
Steps to Use the Formula
- Write the equation in standard form: ax² + bx + c = 0
- Identify a, b, and c
- Calculate the discriminant: b² - 4ac
- Substitute into the formula
- Simplify (two solutions from ± symbol)
Example 1: Two Distinct Roots
Solve: x² - 3x - 10 = 0
Step 1: Identify coefficients
a = 1, b = -3, c = -10
Step 2: Calculate discriminant
Δ = (-3)² - 4(1)(-10) = 9 + 40 = 49
Step 3: Apply formula
x = [3 ± √49] / 2
x = [3 ± 7] / 2
Step 4: Solve
x = (3 + 7)/2 = 5 OR x = (3 - 7)/2 = -2
Solutions: x = 5, x = -2
Example 2: One Repeated Root
Solve: x² - 6x + 9 = 0
a = 1, b = -6, c = 9
Δ = (-6)² - 4(1)(9) = 36 - 36 = 0
x = [6 ± √0] / 2 = 6/2 = 3
Solution: x = 3 (double root)
Example 3: No Real Roots
Solve: x² + 2x + 5 = 0
a = 1, b = 2, c = 5
Δ = 2² - 4(1)(5) = 4 - 20 = -16 < 0
Since discriminant is negative, no real solutions exist.
Complex solutions: x = [-2 ± √(-16)] / 2 = -1 ± 2i
Example 4: Non-Integer Coefficients
Solve: 2x² + 5x - 1 = 0
a = 2, b = 5, c = -1
Δ = 5² - 4(2)(-1) = 25 + 8 = 33
x = [-5 ± √33] / 4
Solutions: x = (-5 + √33)/4 ≈ 0.186
x = (-5 - √33)/4 ≈ -2.686
Method 3: Completing the Square
This method transforms a quadratic into perfect square form, useful for deriving the quadratic formula and graphing.
The Process
- Move the constant term to the right side
- If a ≠ 1, divide everything by a
- Take half of the b coefficient, square it, and add to both sides
- Factor the left side as a perfect square
- Take the square root of both sides
- Solve for x
Example 1: Basic Completing the Square
Solve: x² + 8x + 7 = 0
Step 1: Move constant
x² + 8x = -7
Step 2: Complete the square
Take half of 8: 8/2 = 4
Square it: 4² = 16
Add to both sides: x² + 8x + 16 = -7 + 16
Step 3: Factor left side
(x + 4)² = 9
Step 4: Take square root
x + 4 = ±3
Step 5: Solve
x = -4 + 3 = -1 OR x = -4 - 3 = -7
Solutions: x = -1, x = -7
Example 2: With Leading Coefficient ≠ 1
Solve: 2x² - 12x + 10 = 0
Step 1: Divide by a (2)
x² - 6x + 5 = 0
Step 2: Move constant
x² - 6x = -5
Step 3: Complete the square
(-6/2)² = 9
x² - 6x + 9 = -5 + 9
(x - 3)² = 4
Step 4: Solve
x - 3 = ±2
x = 3 + 2 = 5 OR x = 3 - 2 = 1
Solutions: x = 5, x = 1
Converting to Vertex Form
Completing the square converts y = ax² + bx + c to vertex form: y = a(x - h)² + k
Example: Convert y = x² - 4x + 1 to vertex form
y = x² - 4x + 1
y = (x² - 4x + 4) + 1 - 4
y = (x - 2)² - 3
Vertex form: y = (x - 2)² - 3
Vertex: (2, -3)
Method 4: Graphing
Quadratic equations graph as parabolas. The solutions are where the parabola crosses the x-axis.
Key Features of a Parabola
Vertex: The highest or lowest point
- If a > 0: parabola opens upward (vertex is minimum)
- If a < 0: parabola opens downward (vertex is maximum)
Axis of Symmetry: x = -b/(2a)
Y-intercept: (0, c)
X-intercepts (roots): Solutions to ax² + bx + c = 0
Finding the Vertex
Method 1: Use h = -b/(2a), then find k = f(h)
Method 2: Complete the square to get vertex form y = a(x - h)² + k
Example: Find the vertex of y = 2x² - 8x + 3
Method 1:
h = -(-8)/(2·2) = 8/4 = 2
k = 2(2)² - 8(2) + 3 = 8 - 16 + 3 = -5
Vertex: (2, -5)
Analyzing Solutions from the Graph
- Graph crosses x-axis twice: Two real roots
- Graph touches x-axis once: One repeated root
- Graph doesn't touch x-axis: No real roots
Choosing the Best Method
| Situation | Best Method | |-----------|-------------| | Easy to factor | Factoring | | a = 1, small integers | Factoring | | Not factorable | Quadratic Formula | | Need exact irrational answers | Quadratic Formula | | Finding vertex | Completing the Square | | Deriving formulas | Completing the Square | | Visual understanding | Graphing | | c = 0 | Factor out x first | | b = 0 | Take square root directly |
Real-World Applications
1. Projectile Motion
Problem: A ball is thrown upward with initial velocity 20 m/s from a height of 2 m. When does it hit the ground?
Equation: h(t) = -5t² + 20t + 2 = 0 (where h is height in meters, t is time in seconds)
Using quadratic formula:
a = -5, b = 20, c = 2
t = [-20 ± √(400 + 40)] / (-10)
t = [-20 ± √440] / (-10)
t = [-20 ± 20.98] / (-10)
t = 0.098 s (going up, ignoring)
t = 4.098 s (hitting ground)
Answer: Approximately 4.1 seconds
2. Area Problems
Problem: A rectangular garden has length 4 meters more than its width. If the area is 60 square meters, find the dimensions.
Let width = x
Then length = x + 4
Area: x(x + 4) = 60
x² + 4x = 60
x² + 4x - 60 = 0
Factoring: (x + 10)(x - 6) = 0
x = -10 (reject, negative) or x = 6
Width: 6 meters
Length: 10 meters
3. Business Optimization
Problem: A company's profit P (in thousands) is given by P(x) = -2x² + 40x - 150, where x is units produced (in thousands). Find maximum profit.
Vertex represents maximum (a < 0)
h = -40/(2·(-2)) = 40/4 = 10
k = -2(10)² + 40(10) - 150 = -200 + 400 - 150 = 50
Maximum profit: $50,000 at 10,000 units
4. Engineering and Design
Problem: A bridge arch follows the equation y = -0.01x² + 2x, where y is height in meters. Find the span of the arch (where it meets the ground).
Set y = 0:
-0.01x² + 2x = 0
x(-0.01x + 2) = 0
x = 0 (one end)
-0.01x + 2 = 0
x = 200 (other end)
Span: 200 meters
5. Physics: Acceleration
Problem: A car's position is given by s(t) = 3t² + 2t + 1 meters. When is it at position 50 meters?
3t² + 2t + 1 = 50
3t² + 2t - 49 = 0
Using quadratic formula:
t = [-2 ± √(4 + 588)] / 6
t = [-2 ± √592] / 6
t = [-2 ± 24.33] / 6
t = 3.72 seconds (positive solution)
Sum and Product of Roots
For ax² + bx + c = 0 with roots r₁ and r₂:
Sum of roots: r₁ + r₂ = -b/a
Product of roots: r₁ · r₂ = c/a
Application
Example: Without solving, find the sum and product of roots of 3x² - 12x + 5 = 0
Sum: -(-12)/3 = 12/3 = 4
Product: 5/3
This is useful for:
- Checking solutions
- Creating equations from given roots
- Analyzing equations without solving
Creating Equations from Roots
If you know the roots, you can construct the equation.
Formula: (x - r₁)(x - r₂) = 0
Example: Create a quadratic with roots 3 and -5
(x - 3)(x - (-5)) = 0
(x - 3)(x + 5) = 0
x² + 5x - 3x - 15 = 0
x² + 2x - 15 = 0
Common Mistakes to Avoid
Mistake 1: Forgetting ± in Square Root
Wrong: x² = 9, so x = 3 Right: x² = 9, so x = ±3 (x = 3 or x = -3)
Mistake 2: Sign Errors in Formula
Wrong: For x² - 5x + 6 = 0, using b = 5 Right: b = -5 (include the sign!)
Mistake 3: Dividing by Variable
Wrong: x² = 3x, so x = 3 (dividing both sides by x) Right: x² - 3x = 0, so x(x - 3) = 0, giving x = 0 or x = 3
Mistake 4: Incomplete Factoring
Wrong: 2x² + 4x = 0 → (x + 2)(2x) = 0 Right: 2x² + 4x = 0 → 2x(x + 2) = 0
Mistake 5: Arithmetic Errors
Always double-check:
- Discriminant calculation
- Signs when substituting
- Simplifying radicals
Practice Problems
Beginner:
- Solve: x² + 7x + 12 = 0
- Solve: x² - 16 = 0
- Solve: x² + 4x = 0
Intermediate:
- Solve: 2x² - 5x - 3 = 0
- Solve: x² - 4x + 4 = 0
- Solve: x² + 6x + 2 = 0
Advanced:
- Find the vertex of y = -3x² + 12x - 5
- A rectangle's length is 3 more than twice its width, and area is 44. Find dimensions.
- Solve: (x + 1)² = 5
Answers:
- x = -3, x = -4
- x = 4, x = -4
- x = 0, x = -4
- x = 3, x = -1/2
- x = 2 (double root)
- x = -3 ± √7
- Vertex: (2, 7)
- Width 4, Length 11
- x = -1 ± √5
Advanced Topics
Complex Roots
When Δ < 0, solutions involve imaginary unit i where i² = -1.
Example: x² + 2x + 5 = 0
x = [-2 ± √(4-20)] / 2
x = [-2 ± √(-16)] / 2
x = [-2 ± 4i] / 2
x = -1 ± 2i
Complex roots: -1 + 2i and -1 - 2i (complex conjugates)
Quadratic Inequalities
Solve: x² - 5x + 6 < 0
Step 1: Factor
(x - 2)(x - 3) < 0
Step 2: Find critical points
x = 2, x = 3
Step 3: Test intervals
x < 2: (+)(+) = + (false)
2 < x < 3: (-)(+) = - (true)
x > 3: (+)(+) = + (false)
Solution: 2 < x < 3
Conclusion
Quadratic equations are a cornerstone of algebra with applications across mathematics, science, and engineering. Mastering the various solution methods—factoring, quadratic formula, completing the square, and graphing—gives you flexibility to approach problems efficiently.
Key takeaways:
- Choose the method that best fits the problem
- Always check your solutions by substituting back
- Understand the discriminant to predict the nature of roots
- Recognize real-world applications to motivate learning
- Practice regularly with diverse problems
Whether you're solving for projectile motion, optimizing business scenarios, or analyzing parabolic structures, the skills you've learned here will serve you well. Use our quadratic calculator to practice and verify your solutions as you build confidence in working with quadratic equations.
Remember: every method has its place, and true mastery comes from knowing which tool to use when. Keep practicing, and quadratic equations will become second nature!