The Quadratic Formula

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:

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.