Quadratic Equations#
Linear Equations Revision#
Theory#
Before we dive into the fascinating world of quadratic equations, let’s take a moment to revisit linear equations. Think of this as building a bridge between what you already know and the new concepts we’re about to explore.
Linear Equation Structure: A linear equation has the form:
where \(a \neq 0\). The solution is always \(x = -\frac{b}{a}\).
Key Properties of Linear Equations:
First-degree: The highest power of \(x\) is 1
One solution: Every linear equation has exactly one solution (unless it’s a contradiction or identity)
Straight line graph: When graphed, linear equations form straight lines
Constant rate of change: The relationship between variables is constant
Why This Matters: Linear equations represent situations where change happens at a constant rate. But what happens when change isn’t constant? When acceleration or deceleration occurs? That’s where quadratic equations come in.
Application#
Examples#
Example 1: Simple Linear Equation#
Let’s solve this as a warm-up: \(3x - 12 = 0\)
Method 1: Isolation
\(3x - 12 = 0 \quad \text{(Original equation)}\)
\(3x = 12 \quad \text{(Add 12 to both sides)}\)
\(x = 4 \quad \text{(Divide by 3)}\)
Verification: \(3(4) - 12 = 12 - 12 = 0\) ✓
Example 2: Comparing Linear and Quadratic Behavior#
Notice the difference: Linear equation \(y = 2x + 1\) vs. Quadratic equation \(y = x^2 + 1\)
Linear: As \(x\) increases by 1, \(y\) always increases by 2 (constant change) Quadratic: As \(x\) increases by 1, the change in \(y\) varies (accelerating change)
Interactive Visualization: Linear vs Quadratic Comparison#
Multiple Choice Questions#
Quadratic Equations#
Theory#
Now let’s explore quadratic equations - one of the most beautiful and useful topics in algebra. Here’s why quadratic equations are so important: they describe acceleration, parabolic motion, optimization problems, and countless real-world phenomena where change isn’t constant.
Fundamental Definition: A quadratic equation is a polynomial equation of degree 2, written in standard form as:
where \(a\), \(b\), and \(c\) are real constants with \(a \neq 0\).
Why “Quadratic”?: The term comes from the Latin word “quadratum,” meaning square, because the defining feature is the \(x^2\) term.
Key Characteristics and Properties:
Degree and Solutions: Quadratic equations are second-degree equations, and they can have:
Two distinct real solutions (most common case)
One repeated real solution (when the parabola just touches the x-axis)
Two complex conjugate solutions (when the parabola doesn’t cross the x-axis)
Graphical Representation: The graph of a quadratic function \(y = ax^2 + bx + c\) is always a parabola:
If \(a > 0\): parabola opens upward (U-shaped)
If \(a < 0\): parabola opens downward (∩-shaped)
The Discriminant: The expression \(\Delta = b^2 - 4ac\) determines the nature of solutions:
If \(\Delta > 0\): two distinct real solutions
If \(\Delta = 0\): one repeated real solution
If \(\Delta < 0\): two complex conjugate solutions
Solution Methods - A Complete Toolkit:
1. Factoring Method: When the quadratic can be written as a product of linear factors
Most elegant when it works
Relies on recognizing patterns or systematic trial
Uses the zero product property: if \((p)(q) = 0\), then \(p = 0\) or \(q = 0\)
2. Completing the Square: A systematic method that always works
Transforms the equation into perfect square form
Reveals the vertex of the parabola
Foundation for deriving the quadratic formula
3. Quadratic Formula: The universal solution method $\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)$
Works for any quadratic equation
Provides both solutions simultaneously
Shows the role of the discriminant clearly
4. Graphical Method: Finding x-intercepts of the parabola
Useful for understanding the geometric meaning
Helps visualize the relationship between solutions and graph
Special Quadratic Forms and Patterns:
Perfect Square Trinomials: \(x^2 \pm 2ax + a^2 = (x \pm a)^2\)
Difference of Squares: \(x^2 - a^2 = (x + a)(x - a)\)
Standard Factoring Pattern: \(x^2 + (p + q)x + pq = (x + p)(x + q)\)
Vertex Form: \(y = a(x - h)^2 + k\) where \((h, k)\) is the vertex
Real-World Applications: Quadratic equations model numerous phenomena:
Physics: Projectile motion, falling objects, pendulum motion
Engineering: Bridge design, optimization problems, signal processing
Economics: Profit maximization, cost-revenue analysis
Geometry: Area problems, optimization of shapes
Interactive Visualization: Quadratic Function Explorer#
Application#
Examples#
Example 1: Solving by Factoring#
Let’s solve this step by step: \(x^2 - 5x + 6 = 0\)
Method 1: Factoring Approach
\(x^2 - 5x + 6 = 0 \quad \text{(We need two numbers that multiply to 6 and add to -5)}\)
\((x - 2)(x - 3) = 0 \quad \text{(Since -2 × -3 = 6 and -2 + (-3) = -5)}\)
\(x - 2 = 0 \text{ or } x - 3 = 0 \quad \text{(Zero product property)}\)
\(x = 2 \text{ or } x = 3 \quad \text{(Solutions)}\)
Verification: \(2^2 - 5(2) + 6 = 4 - 10 + 6 = 0\) ✓ and \(3^2 - 5(3) + 6 = 9 - 15 + 6 = 0\) ✓
Example 2: Completing the Square#
Here’s a systematic approach to: \(x^2 + 6x + 5 = 0\)
Method 1: Completing the Square
\(x^2 + 6x + 5 = 0 \quad \text{(Original equation)}\)
\(x^2 + 6x = -5 \quad \text{(Move constant to right side)}\)
\(x^2 + 6x + 9 = -5 + 9 \quad \text{(Add \)(6/2)^2 = 9\( to complete the square)}\)
\((x + 3)^2 = 4 \quad \text{(Factor the perfect square trinomial)}\)
\(x + 3 = \pm 2 \quad \text{(Take square root of both sides)}\)
\(x = -3 + 2 = -1 \text{ or } x = -3 - 2 = -5 \quad \text{(Two solutions)}\)
Method 2: Quadratic Formula Verification
\(x = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2} \quad \text{(Apply formula with a=1, b=6, c=5)}\)
\(x = \frac{-2}{2} = -1 \text{ or } x = \frac{-10}{2} = -5 \quad \text{(Same solutions - good!)}\)
Example 3: Using Multiple Methods#
Let’s solve: \(2x^2 - 7x + 3 = 0\)
Method 1: Factoring (AC Method)
\(2x^2 - 7x + 3 = 0 \quad \text{(Find factors of ac = 6 that add to b = -7)}\)
\(2x^2 - 6x - x + 3 = 0 \quad \text{(Split middle term: -7x = -6x - x)}\)
\(2x(x - 3) - 1(x - 3) = 0 \quad \text{(Factor by grouping)}\)
\((2x - 1)(x - 3) = 0 \quad \text{(Factor out common binomial)}\)
\(x = \frac{1}{2} \text{ or } x = 3 \quad \text{(Solve each factor)}\)
Method 2: Quadratic Formula
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} \quad \text{(Substitute a=2, b=-7, c=3)}\)
\(x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4} \quad \text{(Simplify under the radical)}\)
\(x = \frac{12}{4} = 3 \text{ or } x = \frac{2}{4} = \frac{1}{2} \quad \text{(Same solutions - perfect consistency!)}\)
Multiple Choice Questions#
Sector Specific Questions: Quadratic Equations Applications#
Key Takeaways#
Important
A quadratic equation has the form \(ax^2 + bx + c = 0\) where \(a \neq 0\)
Quadratic equations can have 0, 1, or 2 real solutions depending on the discriminant \(\Delta = b^2 - 4ac\)
Four main solving methods: factoring, completing the square, quadratic formula, and graphing
The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) works for any quadratic equation
The graph of a quadratic function is always a parabola
Vertex occurs at \(x = \frac{-b}{2a}\), which is often the key to optimization problems
Quadratic equations model projectile motion, area optimization, and profit maximization
Always check solutions by substitution, and consider which solutions make sense in context