Quadratic Equation

Lesson Objectives

Understand the general form of a quadratic equation.
Learn how to solve quadratic equations using factoring, completing the square, and the quadratic formula.
Explore the discriminant and its role in determining the nature of the roots.
Apply quadratic equations to solve real-world problems.

Lesson Introduction

Imagine you're designing a parabolic bridge arch, where the shape of the arch follows a quadratic equation. In this lesson, you will learn how to solve quadratic equations, which will allow you to calculate the height of the arch at any point along its span. This is just one example of how quadratic equations are used in real-life applications like physics, engineering, and architecture.

Core Lesson Content

A quadratic equation is any equation that can be written in the form: $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants, and $a \neq 0$ .
The solutions to a quadratic equation are known as the "roots" or "zeroes" of the equation. These can be found using various methods, such as:

Factoring: When the equation can be factored into two binomials.
Completing the Square: A method where we manipulate the equation to form a perfect square trinomial.
The Quadratic Formula: A formula used when the equation cannot be easily factored. The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where the discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots.

Worked Examples

Example 1 (Basic):
Solve

x^2 - 5x + 6 = 0

by factoring.
Factor the equation:

(x - 2)(x - 3) = 0

Set each factor equal to zero:

x - 2 = 0 \quad \text{or} \quad x - 3 = 0

Solve for

x

x = 2 \quad \text{or} \quad x = 3

Example 2 (Intermediate):
Solve

x^2 + 4x + 3 = 0

using the quadratic formula.
The quadratic formula is

x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)}

Simplifying:

x = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2}

So,

x = -1

x = -3

Example 3 (Advanced):
Solve

2x^2 + 3x - 2 = 0

by completing the square.
Divide by 2:

x^2 + \frac{3}{2}x - 1 = 0

Move the constant to the other side:

x^2 + \frac{3}{2}x = 1

Complete the square:

x^2 + \frac{3}{2}x + \left(\frac{3}{4}\right)^2 = 1 + \left(\frac{3}{4}\right)^2

\Rightarrow \left(x + \frac{3}{4}\right)^2 = \frac{19}{16}

Taking the square root:

x + \frac{3}{4} = \pm \frac{\sqrt{19}}{4}

Finally:

x = \frac{-3 \pm \sqrt{19}}{4}

Exercises

Solve $x^2 - 7x + 12 = 0$ by factoring.
Use the quadratic formula to solve $2x^2 + 5x - 3 = 0$ .
Find the roots of $x^2 + 6x + 9 = 0$ by completing the square.
Solve the quadratic equation $x^2 - 4x - 5 = 0$ by factoring.
[WAEC] Solve $x^2 + 7x + 10 = 0$ using the quadratic formula. (Past Question)
[NECO] Solve $3x^2 - 2x - 5 = 0$ by completing the square. (Past Question)
[JAMB] Solve $x^2 - 9x + 20 = 0$ by factoring. (Past Question)
Solve $x^2 + 2x - 8 = 0$ using the quadratic formula.
[NABTEC] Solve $4x^2 - 12x + 9 = 0$ by completing the square. (Past Question)
Solve $x^2 + 5x + 4 = 0$ by factoring.

Conclusion / Recap

In this lesson, you've learned how to solve quadratic equations using factoring, completing the square, and the quadratic formula. You've also explored the discriminant and its role in determining the nature of the roots.
Next up: Linear Programming and Optimization — applying quadratic functions to solve real-world optimization problems!