# The Quadratic Formula

Imagine you're designing a water fountain and need to know where the water arc will hit the ground, or calculating the trajectory of a basketball to determine if it will go through the hoop. These situations involve quadratic equations, and the quadratic formula is your universal tool for solving them.

## What Is a Quadratic Equation?

A quadratic equation is any equation that can be written in the form:

$$ax^2 + bx + c = 0$$

where: - $a$, $b$, and $c$ are real numbers (called coefficients) - $a \neq 0$ (if $a = 0$, it's not quadratic, just linear) - $x$ is the variable we're solving for

The solutions (also called roots or zeros) are the values of $x$ that make the equation true.

## The Quadratic Formula

The quadratic formula gives us the solutions to any quadratic equation $ax^2 + bx + c = 0$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The symbol $\pm$ ("plus or minus") means there are typically two solutions: - $x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ - $x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

## Why Does the Quadratic Formula Work?

The quadratic formula comes from a method called completing the square. Here's a simplified version of the derivation:

Starting with $ax^2 + bx + c = 0$:

1. Divide everything by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ 2. Move the constant: $x^2 + \frac{b}{a}x = -\frac{c}{a}$ 3. Complete the square by adding $(\frac{b}{2a})^2$ to both sides 4. Factor the left side as a perfect square 5. Take the square root of both sides 6. Solve for $x$

The result is the quadratic formula. This proves it works for every quadratic equation!

## When Should You Use the Quadratic Formula?

You have several methods for solving quadratic equations. Here's when to use each:

Use the Quadratic Formula when: - The equation doesn't factor easily - The coefficients are large or complicated - You need exact answers with irrational numbers - You're not sure which method will work - You want a reliable method that always works

Consider other methods when: - Factoring: When the equation factors nicely (like $x^2 - 5x + 6 = 0$) - Square Root Method: When there's no $x$ term (like $x^2 = 25$ or $(x-3)^2 = 16$) - Completing the Square: When the leading coefficient is 1 and $b$ is even

## The Discriminant: Predicting Your Solutions

The expression under the square root, $b^2 - 4ac$, is called the discriminant. It tells you what kind of solutions to expect before you solve:

| Discriminant | Number of Solutions | Type of Solutions | |--------------|---------------------|-------------------| | $b^2 - 4ac > 0$ | Two different solutions | Real numbers | | $b^2 - 4ac = 0$ | One repeated solution | Real number | | $b^2 - 4ac < 0$ | No real solutions | Complex numbers |

## Example 1: Solving a Quadratic Equation

Problem: Solve $2x^2 + 7x - 15 = 0$

Step 1: Identify a, b, and c

Comparing to the standard form $ax^2 + bx + c = 0$: - $a = 2$ - $b = 7$ - $c = -15$

Step 2: Calculate the Discriminant

$$b^2 - 4ac = (7)^2 - 4(2)(-15)$$ $$= 49 - (-120)$$ $$= 49 + 120 = 169$$

Since $169 > 0$, we expect two real solutions.

Step 3: Apply the Quadratic Formula

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{169}}{2(2)}$$

$$x = \frac{-7 \pm 13}{4}$$

Step 4: Find Both Solutions

$$x_1 = \frac{-7 + 13}{4} = \frac{6}{4} = \frac{3}{2}$$

$$x_2 = \frac{-7 - 13}{4} = \frac{-20}{4} = -5$$

Solutions: $x = \frac{3}{2}$ or $x = -5$

## Example 2: An Equation with Irrational Solutions

Problem: Solve $x^2 - 6x + 2 = 0$

Step 1: Identify a, b, and c - $a = 1$ - $b = -6$ - $c = 2$

Step 2: Calculate the Discriminant

$$b^2 - 4ac = (-6)^2 - 4(1)(2) = 36 - 8 = 28$$

Since $28 > 0$, we expect two real solutions. Since 28 is not a perfect square, they will be irrational numbers.

Step 3: Apply the Quadratic Formula

$$x = \frac{-(-6) \pm \sqrt{28}}{2(1)} = \frac{6 \pm \sqrt{28}}{2}$$

Step 4: Simplify the Square Root

$\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}$

$$x = \frac{6 \pm 2\sqrt{7}}{2} = \frac{6}{2} \pm \frac{2\sqrt{7}}{2} = 3 \pm \sqrt{7}$$

Solutions: $x = 3 + \sqrt{7} \approx 5.65$ or $x = 3 - \sqrt{7} \approx 0.35$

## Step-by-Step Summary

To solve a quadratic equation using the quadratic formula:

1. Write in standard form: Get $ax^2 + bx + c = 0$ 2. Identify coefficients: Find values of $a$, $b$, and $c$ 3. Calculate discriminant: Find $b^2 - 4ac$ to predict solution type 4. Substitute into formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ 5. Simplify: Calculate both solutions (+ and -) 6. Check: Substitute solutions back into original equation

## Real-World Applications

The quadratic formula appears in many real-world situations:

- Physics: Projectile motion (Where does a ball land?) - Engineering: Bridge and arch design - Business: Profit optimization (maximizing revenue) - Architecture: Calculating dimensions for parabolic structures - Sports: Analyzing ball trajectories

Whenever a relationship involves squared quantities, quadratic equations and the quadratic formula become essential tools.