Quadratic Equations, Functions, Zeros, and Models (Section 3.2 Study Notes)

Quadratic Equations and Functions

Introduction

This section covers the fundamental concepts of quadratic equations and functions, including methods for finding their zeros, solving quadratic equations, and applying these techniques to real-world problems. Quadratic equations and functions are central topics in precalculus and serve as a foundation for further study in mathematics.

Quadratic Equations

• Definition: A quadratic equation is any equation that can be written in the form: $ax^2 + bx + c = 0$, where $a \neq 0$ and $a$, $b$, $c$ are real numbers.

• Standard Form: The above form is called the standard form of a quadratic equation.

Quadratic Functions

• Definition: A quadratic function $f$ is a function that can be written as: $f(x) = ax^2 + bx + c$, where $a \neq 0$ and $a$, $b$, $c$ are real numbers.

• Zeros of a Quadratic Function: The zeros of $f(x)$ are the solutions to $ax^2 + bx + c = 0$. These are also called roots or solutions of the equation.

• Quadratic functions and equations can have real-number zeros or imaginary-number zeros (complex roots).

Equation-Solving Principles

Principle of Zero Products

• If $ab = 0$, then $a = 0$ or $b = 0$.

• Conversely, if $a = 0$ or $b = 0$, then $ab = 0$.

• This principle is used to solve equations that have been factored.

Principle of Square Roots

• If $x^2 = k$, then $x = \sqrt{k}$ or $x = -\sqrt{k}$.

• This principle is used when the quadratic equation can be rewritten so that the variable term is squared and isolated.

Methods for Solving Quadratic Equations

Factoring

• Rewrite the equation in standard form: $ax^2 + bx + c = 0$.

• Factor the quadratic expression, if possible.

• Set each factor equal to zero and solve for $x$ using the principle of zero products.

• Example: Solve $2x^2 - x = 3$. $2x^2 - x - 3 = 0$ $(x + 1)(2x - 3) = 0$ $x + 1 = 0$ or $2x - 3 = 0$ $x = -1$ or $x = \frac{3}{2}$

Using the Principle of Square Roots

• Isolate the squared term and solve for $x$ using square roots.

• Example: Solve $2x^2 - 10 = 0$. $2x^2 = 10$ $x^2 = 5$ $x = \sqrt{5}$ or $x = -\sqrt{5}$

Completing the Square

• Move variable terms to one side and arrange in descending order.

• Divide by the coefficient of $x^2$ if it is not 1.

• Find half the coefficient of $x$, square it, and add to both sides.

• Express one side as a perfect square trinomial.

• Use the principle of square roots to solve for $x$.

• Example: Solve $2x^2 - 1 = 3x$. $2x^2 - 3x - 1 = 0$ $x^2 - \frac{3}{2}x = \frac{1}{2}$ Add $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$ to both sides: $x^2 - \frac{3}{2}x + \frac{9}{16} = \frac{1}{2} + \frac{9}{16}$ $\left(x - \frac{3}{4}\right)^2 = \frac{17}{16}$ $x - \frac{3}{4} = \pm \frac{\sqrt{17}}{4}$ $x = \frac{3 \pm \sqrt{17}}{4}$

Quadratic Formula

• The quadratic formula gives the solutions to $ax^2 + bx + c = 0$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• This formula can be used for any quadratic equation, regardless of whether it can be factored.

• Example: Solve $3x^2 + 2x = 7$. $3x^2 + 2x - 7 = 0$ $a = 3$, $b = 2$, $c = -7$ $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-7)}}{2 \cdot 3}$ $x = \frac{-2 \pm \sqrt{4 + 84}}{6}$ $x = \frac{-2 \pm \sqrt{88}}{6}$ $x \approx -1.897$ or $x \approx 1.230$

Discriminant

• The discriminant is the expression $b^2 - 4ac$ in the quadratic formula.

• It determines the nature of the solutions:

  • $b^2 - 4ac = 0$ → One real-number solution

  • $b^2 - 4ac > 0$ → Two distinct real-number solutions

  • $b^2 - 4ac < 0$ → Two complex (imaginary) solutions

Equations Reducible to Quadratic Form

Introduction

Some equations can be transformed into quadratic form by substitution, allowing them to be solved using quadratic techniques.

• Method: Substitute $u = x^2$ (or another appropriate expression) to rewrite the equation as a quadratic in $u$.

• Solve for $u$ using factoring or the quadratic formula, then substitute back and solve for $x$.

• Example: Solve $x^4 - 5x^2 + 4 = 0$. Let $u = x^2$, so $u^2 - 5u + 4 = 0$ $(u - 1)(u - 4) = 0$ $u = 1$ or $u = 4$ $x^2 = 1$ or $x^2 = 4$ $x = \pm 1$ or $x = \pm 2$

Applications of Quadratic Equations

Introduction

Quadratic equations are used to model and solve real-world problems in various fields, such as business, physics, and engineering.

• Example: Magazine Closures The function $m(x) = 34x^2 - 59x + 81$ estimates the number of magazine closures after 2012. To find the year when closures reached 99: Set $m(x) = 99$: $99 = 34x^2 - 59x + 81$ $0 = 34x^2 - 59x - 18$ Use the quadratic formula with $a = 34$, $b = -59$, $c = -18$: $x = \frac{59 \pm \sqrt{(-59)^2 - 4 \cdot 34 \cdot (-18)}}{2 \cdot 34}$ $x \approx 2$ or $x \approx -0.3$ Since $x$ must be positive (years after 2012), the answer is $x = 2$, so the year is 2014.

Summary Table: Methods for Solving Quadratic Equations

Method	When to Use	Example
Factoring	When the quadratic can be factored easily	$x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0$
Square Roots	When the equation is in the form $x^2 = k$	$x^2 = 9$ $x = \pm 3$
Completing the Square	When $a = 1$ or can be made 1, and factoring is difficult	$x^2 + 6x + 5 = 0$ Complete the square
Quadratic Formula	Always works for any quadratic equation	$x^2 + 4x + 1 = 0$ $x = \frac{-4 \pm \sqrt{16 - 4}}{2}$

Additional info: These notes expand on the provided slides and handwritten content, adding definitions, step-by-step methods, and a summary table for clarity and completeness.