Quadratic Functions V2#
Linear Equations Revision#
Theory#
Before we dive into the fascinating world of quadratic functions, let’s take a moment to revisit our foundation - linear equations. This revision is crucial because every skill you’ve mastered with linear equations becomes a stepping stone for understanding quadratics!
A linear equation in standard form is:
where \(a \neq 0\). Here’s why this matters: the systematic approach we use to solve linear equations - isolating the variable, maintaining balance, and step-by-step manipulation - forms the foundation for all algebraic problem-solving.
The Solution Process:
This gives us the solution \(x = -\frac{b}{a}\), which represents where the line \(y = ax + b\) crosses the x-axis.
Tip
When solving linear equations, always perform the same operation to both sides of the equation. This fundamental principle of maintaining equality will be essential when we tackle quadratic equations!
Interactive Visualization: Linear Equations#
Application#
Examples#
Example 1: Building Confidence with Linear Equations#
Let’s solve: \(3x - 12 = 0\)
Method 1: Direct Manipulation
\(3x - 12 = 0 \quad \text{(starting equation)}\)
\(3x = 12 \quad \text{(adding 12 to both sides)}\)
\(x = 4 \quad \text{(dividing both sides by 3)}\)
Method 2: Formula Application
With \(a = 3\) and \(b = -12\):
\(x = -\frac{b}{a} = -\frac{(-12)}{3} = \frac{12}{3} = 4 \quad \text{(same answer!)}\)
Example 2: Fractional Coefficients#
Let’s work through: \(\frac{2}{3}x + 4 = 0\)
Method 1: Clear Fractions First
\(\frac{2}{3}x + 4 = 0 \quad \text{(multiply everything by 3)}\)
\(2x + 12 = 0 \quad \text{(now we have integers!)}\)
\(2x = -12 \quad \text{(subtract 12 from both sides)}\)
\(x = -6 \quad \text{(divide by 2)}\)
Multiple Choice Questions#
Sector Specific Questions: Linear Equations Applications#
Warning
Don’t rush through linear equation review! The systematic thinking you develop here - isolating variables, maintaining equation balance, checking solutions - is exactly what you’ll need for quadratic equations.
Key Takeaways#
Important
Linear equations have the form \(ax + b = 0\) and represent straight lines when graphed
Systematic solution process: Isolate the variable by performing inverse operations in the correct order
Balance principle: Whatever operation you do to one side, you must do to the other side
Solution verification: Always substitute your answer back into the original equation to check
Foundation skills: These algebraic manipulation techniques are essential for solving quadratic equations
Quadratic Equations and their Form#
Theory#
Now here’s where mathematics becomes truly exciting! A quadratic equation is our gateway from the straight-line world of linear equations into the curved, elegant world of parabolas. Let’s explore what makes these equations so special and powerful.
The Standard Form of a Quadratic Equation
A quadratic equation in standard form is:
where \(a\), \(b\), and \(c\) are constants, and crucially, \(a \neq 0\). Notice what happens if \(a = 0\) - we’d just have \(bx + c = 0\), which is a linear equation!
Understanding the Components
The \(ax^2\) term: This is what makes it quadratic! The presence of \(x^2\) creates the characteristic curved shape when graphed.
The \(bx\) term: This linear component influences the position and orientation of the parabola.
The \(c\) term: This constant determines where the parabola crosses the y-axis.
Important
The coefficient \(a\) determines the direction of the parabola: if \(a > 0\), it opens upward (like a smile); if \(a < 0\), it opens downward (like a frown).
From Quadratic Functions to Quadratic Equations
When we have a quadratic function \(f(x) = ax^2 + bx + c\) and we want to find where it crosses the x-axis, we set \(f(x) = 0\), giving us the quadratic equation \(ax^2 + bx + c = 0\).
Recognizing Quadratic Equations
Not all equations with \(x^2\) terms are quadratic equations! Here’s how to identify them:
✓ True quadratic equations:
\(x^2 + 3x - 4 = 0\) (standard form)
\(2x^2 = 8\) (missing linear term, but still quadratic)
\((x-1)(x+3) = 0\) (factored form)
✗ Not quadratic equations:
\(x^3 + x^2 - 2 = 0\) (highest power is 3, so it’s cubic)
\(\frac{1}{x^2} + x = 0\) (negative power of x)
Tip
To identify a quadratic equation, first expand and simplify it completely. If the highest power of \(x\) is 2, and the coefficient of \(x^2\) is not zero, then it’s quadratic!
Interactive Visualization: Quadratic Equation Form#
Application#
Examples#
Example 1: Converting to Standard Form#
Let’s work through: \((2x - 1)(x + 3) = 5\)
Method 1: Expand then Rearrange
\((2x - 1)(x + 3) = 5 \quad \text{(expand the left side)}\)
\(2x^2 + 6x - x - 3 = 5 \quad \text{(using FOIL method)}\)
\(2x^2 + 5x - 3 = 5 \quad \text{(combining like terms)}\)
\(2x^2 + 5x - 8 = 0 \quad \text{(subtracting 5 from both sides)}\)
Now we have it in standard form with \(a = 2\), \(b = 5\), \(c = -8\).
Example 2: Identifying Quadratic Equations#
Let’s solve: Is \(3x^2 + \frac{2}{x} - 1 = 0\) a quadratic equation?
Method 1: Analyze the Terms
\(3x^2 + \frac{2}{x} - 1 = 0 \quad \text{(rewrite the fraction)}\)
\(3x^2 + 2x^{-1} - 1 = 0 \quad \text{(now we can see the powers clearly)}\)
This equation contains \(x^{-1}\) (a negative power), so it’s not a quadratic equation. A true quadratic can only have non-negative integer powers of \(x\), with the highest power being 2.
Multiple Choice Questions#
Sector Specific Questions: Quadratic Form Applications#
Note
The standard form \(ax^2 + bx + c = 0\) is more than just a format - it’s the launching pad for all quadratic solution methods. Getting comfortable with converting equations to this form will save you time and reduce errors throughout this topic.
Key Takeaways#
Important
Standard form is \(ax^2 + bx + c = 0\) where \(a ≠ 0\) is the essential requirement
Coefficient significance: \(a\) determines parabola direction, \(b\) affects position, \(c\) is the y-intercept
Conversion skills: Always expand, simplify, and rearrange to get equations in standard form
Recognition patterns: Look for \(x^2\) as the highest power with integer exponents only
Foundation for solutions: Standard form enables all quadratic solving methods
Real-world appearance: Quadratic equations naturally model projectile motion, optimization, and curved relationships
Factorizing Quadratic Equations by Inspection#
Theory#
Now let’s explore one of the most elegant methods for solving quadratic equations - factorization by inspection! This method is like finding the perfect key for a lock - when it works, it’s incredibly satisfying and efficient.
The Foundation: Zero Product Property
The power of factorization lies in a fundamental property: if \(A \times B = 0\), then either \(A = 0\) or \(B = 0\) (or both). This means if we can write our quadratic as a product of two factors equal to zero, we can solve it immediately!
When \(a = 1\): The Simpler Case
For equations like \(x^2 + bx + c = 0\), we look for two numbers that:
Multiply to give \(c\) (the constant term)
Add to give \(b\) (the coefficient of \(x\))
If we find such numbers \(p\) and \(q\), then: $\(x^2 + bx + c = (x + p)(x + q)\)$
When \(a ≠ 1\): The Challenging Case
For equations like \(ax^2 + bx + c = 0\) where \(a ≠ 1\), we look for two numbers that:
Multiply to give \(ac\) (product of first and last coefficients)
Add to give \(b\) (the middle coefficient)
Tip
Start with factorization when you see “nice” integer coefficients, especially when \(a = 1\). This method is often the fastest when it works!
Strategic Approach
Check for common factors first - factor out any common terms
Identify the type - is \(a = 1\) or \(a ≠ 1\)?
Find the number pair using the appropriate rule
Write the factors and solve using the zero product property
Always verify your solutions in the original equation
Warning
Factorization doesn’t always work with integer factors! If you can’t find suitable integers quickly, move to completing the square or the quadratic formula.
Application#
Examples#
Example 1: Factorization when \(a = 1\)#
Let’s solve: \(x^2 + 7x + 12 = 0\)
Method 1: Find the Number Pair
We need two numbers that multiply to 12 and add to 7.
\(x^2 + 7x + 12 = 0 \quad \text{(factors of 12: 1×12, 2×6, 3×4)}\)
\(x^2 + 7x + 12 = 0 \quad \text{(which pair adds to 7? That's 3 and 4!)}\)
\((x + 3)(x + 4) = 0 \quad \text{(factored form)}\)
\(x + 3 = 0 \text{ or } x + 4 = 0 \quad \text{(zero product property)}\)
\(x = -3 \text{ or } x = -4 \quad \text{(our solutions)}\)
Example 2: Factorization when \(a ≠ 1\)#
Let’s work through: \(6x^2 + 7x + 2 = 0\)
Method 1: Find Factors of \(ac\)
We need two numbers that multiply to \(ac = 6 × 2 = 12\) and add to \(b = 7\).
\(6x^2 + 7x + 2 = 0 \quad \text{(factors of 12: 1×12, 2×6, 3×4)}\)
\(6x^2 + 7x + 2 = 0 \quad \text{(which pair adds to 7? That's 3 and 4!)}\)
\(6x^2 + 3x + 4x + 2 = 0 \quad \text{(split the middle term)}\)
\(3x(2x + 1) + 2(2x + 1) = 0 \quad \text{(factor by grouping)}\)
\((3x + 2)(2x + 1) = 0 \quad \text{(factored form)}\)
\(3x + 2 = 0 \text{ or } 2x + 1 = 0 \quad \text{(zero product property)}\)
\(x = -\frac{2}{3} \text{ or } x = -\frac{1}{2} \quad \text{(our solutions)}\)
Example 3: When Factorization Doesn’t Work Nicely#
Let’s solve: \(x^2 + 3x + 1 = 0\)
Method 1: Attempt Factorization
We need two numbers that multiply to 1 and add to 3.
\(x^2 + 3x + 1 = 0 \quad \text{(factors of 1: only 1×1)}\)
\(x^2 + 3x + 1 = 0 \quad \text{(but 1 + 1 = 2, not 3!)}\)
Since we can’t find integer factors, this equation requires the quadratic formula or completing the square. This shows that factorization has limitations!
Multiple Choice Questions#
Sector Specific Questions: Factorization Applications#
See also
Factorization connects beautifully to polynomial theory and will reappear when you study cubic equations, rational functions, and even complex numbers. The zero product property is fundamental across all of algebra!
Key Takeaways#
Important
Zero product property is the foundation: if \(A × B = 0\), then \(A = 0\) or \(B = 0\)
When \(a = 1\): Find two numbers that multiply to \(c\) and add to \(b\)
When \(a ≠ 1\): Find two numbers that multiply to \(ac\) and add to \(b\), then use grouping
Efficiency advantage: Factorization is the fastest method when it works with integer factors
Limitation awareness: Not all quadratics factor nicely - be ready to switch methods
Always verify: Check your solutions by substituting back into the original equation
Deriving and Using the Quadratic Formula#
Theory#
Welcome to one of mathematics’ most powerful and reliable tools - the quadratic formula! This beautiful equation can solve ANY quadratic equation, making it the “universal key” for quadratic problems. Let’s explore how this formula is born from completing the square and why it’s so incredibly useful.
The Journey to the Formula
The quadratic formula doesn’t just appear from nowhere - it’s derived by completing the square on the general form \(ax^2 + bx + c = 0\). Let’s see this elegant derivation:
Starting with: \(ax^2 + bx + c = 0\)
\(ax^2 + bx = -c \quad \text{(moving the constant)}\)
\(x^2 + \frac{b}{a}x = -\frac{c}{a} \quad \text{(dividing by a)}\)
Now we complete the square by adding \(\left(\frac{b}{2a}\right)^2\) to both sides:
\(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2\)
\(\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2} \quad \text{(perfect square on left)}\)
\(\left(x + \frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}\)
Taking square roots of both sides:
\(x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}\)
\(x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}\)
The Quadratic Formula:
Important
The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) works for ANY quadratic equation in the form \(ax^2 + bx + c = 0\) where \(a ≠ 0\).
Understanding the Components
\(-b\): This centers the formula around the axis of symmetry
\(\pm\): This gives us both roots (the parabola typically crosses the x-axis at two points)
\(b^2 - 4ac\): This is the discriminant - it determines the nature of the roots
\(2a\): This accounts for the parabola’s width and orientation
When to Use the Quadratic Formula
✓ Always works - unlike factorization, which only works for some equations
✓ Decimal coefficients - handles non-integer coefficients perfectly
✓ Irrational solutions - gives exact answers using radicals
✓ Complex solutions - works even when solutions aren’t real numbers
Tip
Memorize the quadratic formula! It’s worth investing time to know it by heart: “x equals negative b plus or minus the square root of b squared minus four a c, all over two a.”
Interactive Visualization: Quadratic Formula Explorer#
Application#
Examples#
Example 1: Standard Application#
Let’s solve: \(2x^2 + 5x - 3 = 0\)
Method 1: Apply the Quadratic Formula
We identify: \(a = 2\), \(b = 5\), \(c = -3\)
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{(substitute our values)}\)
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} \quad \text{(calculate discriminant)}\)
\(x = \frac{-5 \pm \sqrt{25 + 24}}{4} \quad \text{(simplify under radical)}\)
\(x = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4} \quad \text{(perfect square!)}\)
\(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \text{ or } x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)
Example 2: Irrational Solutions#
Let’s work through: \(x^2 + 4x + 1 = 0\)
Method 1: Quadratic Formula with Irrational Result
We identify: \(a = 1\), \(b = 4\), \(c = 1\)
\(x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} \quad \text{(substitute values)}\)
\(x = \frac{-4 \pm \sqrt{16 - 4}}{2} \quad \text{(calculate discriminant)}\)
\(x = \frac{-4 \pm \sqrt{12}}{2} \quad \text{(simplify radical)}\)
\(x = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \quad \text{(exact solutions)}\)
So \(x = -2 + \sqrt{3} \approx -0.27\) or \(x = -2 - \sqrt{3} \approx -3.73\)
Example 3: No Real Solutions#
Let’s solve: \(x^2 + 2x + 5 = 0\)
Method 1: Recognizing Complex Solutions
We identify: \(a = 1\), \(b = 2\), \(c = 5\)
\(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2(1)} \quad \text{(substitute values)}\)
\(x = \frac{-2 \pm \sqrt{4 - 20}}{2} \quad \text{(calculate discriminant)}\)
\(x = \frac{-2 \pm \sqrt{-16}}{2} \quad \text{(negative under radical!)}\)
Since we have \(\sqrt{-16}\), this equation has no real solutions. The graph of \(y = x^2 + 2x + 5\) never crosses the x-axis.
Multiple Choice Questions#
Sector Specific Questions: Quadratic Formula Applications#
Note
The quadratic formula connects beautifully to many advanced topics: it leads naturally to complex numbers, connects to the discriminant (our next topic), and forms the foundation for understanding polynomial equations in general.
Key Takeaways#
Important
Universal solver: The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) works for ANY quadratic equation
Derived from completing the square: Understanding the derivation helps you remember and trust the formula
Handles all cases: Works when factorization fails, especially with decimal coefficients or irrational solutions
Exact answers: Gives precise results using radicals rather than decimal approximations
Memorization essential: This formula is worth committing to memory for lifelong use
Real-world power: Indispensable for engineering, science, and complex problem-solving where “nice” integer solutions are rare
The Discriminant and Nature of Roots#
Theory#
Now let’s explore the discriminant - a powerful mathematical detective that can tell us everything about a quadratic equation’s solutions before we even solve it! The discriminant is the expression under the square root in the quadratic formula, and it holds the key to understanding the nature of solutions.
The Discriminant Formula
For any quadratic equation \(ax^2 + bx + c = 0\), the discriminant is:
(The symbol \(\Delta\) is the Greek letter “delta,” commonly used for the discriminant.)
The Three Cases: A Complete Classification
The discriminant completely determines what kinds of solutions we’ll get:
Case 1: \(\Delta > 0\) (Positive Discriminant)
Two distinct real roots
The parabola crosses the x-axis at two different points
We get two different rational or irrational solutions
Example: \(x^2 - 5x + 6 = 0\) has \(\Delta = 25 - 24 = 1 > 0\)
Case 2: \(\Delta = 0\) (Zero Discriminant)
One repeated real root (also called a “double root”)
The parabola touches the x-axis at exactly one point (the vertex)
We get one solution that appears twice
Example: \(x^2 - 4x + 4 = 0\) has \(\Delta = 16 - 16 = 0\)
Case 3: \(\Delta < 0\) (Negative Discriminant)
No real roots (two complex conjugate roots)
The parabola doesn’t touch the x-axis at all
We get complex solutions involving \(i = \sqrt{-1}\)
Example: \(x^2 + x + 1 = 0\) has \(\Delta = 1 - 4 = -3 < 0\)
Important
The discriminant \(\Delta = b^2 - 4ac\) is like a mathematical crystal ball - it predicts the nature of solutions before you solve the equation!
Special Case: Perfect Square Discriminants
When \(\Delta\) is a perfect square (like 0, 1, 4, 9, 16, …), the quadratic equation has rational solutions. This often means the equation can be factored with integer coefficients!
Geometric Interpretation
The discriminant directly relates to the graph of \(y = ax^2 + bx + c\):
\(\Delta > 0\): Graph crosses x-axis twice
\(\Delta = 0\): Graph touches x-axis once (vertex on x-axis)
\(\Delta < 0\): Graph doesn’t touch x-axis
Tip
Calculate the discriminant first when analyzing quadratic equations! It immediately tells you what type of solutions to expect and which solution method might be most efficient.
Connection to Other Concepts
The discriminant connects to several important ideas:
Factorization: If \(\Delta\) is a perfect square, factorization often works
Vertex: When \(\Delta = 0\), the vertex lies on the x-axis
Optimization: In real-world problems, \(\Delta = 0\) often represents optimal conditions
Complex numbers: Negative discriminants introduce complex solutions
Warning
Be careful with signs when calculating \(b^2 - 4ac\)! The most common error is incorrectly handling negative values of \(b\) or \(c\).
Application#
Examples#
Example 1: Analyzing Without Solving#
Let’s analyze: \(2x^2 - 7x + 3 = 0\)
Method 1: Discriminant Analysis
We identify: \(a = 2\), \(b = -7\), \(c = 3\)
\(\Delta = b^2 - 4ac \quad \text{(apply discriminant formula)}\)
\(\Delta = (-7)^2 - 4(2)(3) \quad \text{(substitute values carefully)}\)
\(\Delta = 49 - 24 = 25 \quad \text{(calculate)}\)
Since \(\Delta = 25 > 0\) and \(25 = 5^2\) (perfect square), this equation has two distinct rational roots. We know without solving that factorization will work nicely!
Example 2: Vertex on x-axis#
Let’s work through: \(x^2 + 6x + 9 = 0\)
Method 1: Recognizing Special Cases
We identify: \(a = 1\), \(b = 6\), \(c = 9\)
\(\Delta = 6^2 - 4(1)(9) \quad \text{(calculate discriminant)}\)
\(\Delta = 36 - 36 = 0 \quad \text{(zero discriminant!)}\)
Since \(\Delta = 0\), this equation has one repeated real root. The parabola \(y = x^2 + 6x + 9\) touches the x-axis at exactly one point. Notice that this factors as \((x + 3)^2 = 0\), giving \(x = -3\) twice.
Example 3: No Real Solutions#
Let’s solve: \(3x^2 + 2x + 5 = 0\)
Method 1: Discriminant First
We identify: \(a = 3\), \(b = 2\), \(c = 5\)
\(\Delta = 2^2 - 4(3)(5) \quad \text{(calculate discriminant)}\)
\(\Delta = 4 - 60 = -56 \quad \text{(negative discriminant)}\)
Since \(\Delta = -56 < 0\), this equation has no real solutions. The parabola \(y = 3x^2 + 2x + 5\) never crosses the x-axis. In the complex number system, the solutions would involve \(\sqrt{-56} = 2i\sqrt{14}\).
Multiple Choice Questions#
Sector Specific Questions: Discriminant Applications#
See also
The discriminant appears throughout advanced mathematics: in conic sections (distinguishing parabolas, ellipses, and hyperbolas), in the study of polynomial equations of higher degree, and in numerical analysis for understanding solution sensitivity.
Key Takeaways#
Important
Discriminant formula: \(\Delta = b^2 - 4ac\) predicts solution types before solving
Three cases: \(\Delta > 0\) (two real roots), \(\Delta = 0\) (one repeated root), \(\Delta < 0\) (no real roots)
Perfect squares: When \(\Delta\) is a perfect square, solutions are rational and factorization often works
Geometric meaning: Discriminant determines how many times the parabola crosses the x-axis
Problem-solving efficiency: Calculate discriminant first to choose the best solution method
Real-world analysis: Use discriminant to analyze system behavior, stability, and feasibility before detailed calculations