Quadratic Equations II

Grade 12 Math - Advanced Problem Solving with Quadratic Equations

Lesson Objectives

Apply factorization, completing the square, and the quadratic formula to solve complex quadratic equations.
Solve real-world word problems that lead to quadratic equations.
Handle quadratic equations embedded in rational, geometric, or contextual settings.

Lesson Introduction

Quadratic equations are equations of the form $ax^2 + bx + c = 0$ , where $a \ne 0$ . In this lesson, we focus on challenging word problems and complex algebraic structures that lead to quadratic equations. Solving such equations requires a deep understanding of algebraic manipulation and conceptual reasoning.

Core Lesson Content

Worked Example

Solving Quadratic Equations

Recall the standard solution methods:

Factorization
Completing the square
Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example 1: Solve

x^2 - 5x + 6 = 0

by factorization.

(x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3

Example 2: Solve

2x^2 + 3x - 2 = 0

using the quadratic formula.

a = 2, b = 3, c = -2

x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4}

x = \frac{-3 \pm 5}{4} \Rightarrow x = \frac{1}{2}, x = -2

Advanced Word Problems Leading to Quadratic Equations

Example 3: The product of two consecutive integers is 132. Find the integers.
Let the first number be

x

, the next is

x + 1

x(x + 1) = 132 \Rightarrow x^2 + x - 132 = 0

Factor:

(x - 11)(x + 12) = 0 \Rightarrow x = 11 \text{ or } x = -12

So the integers are either 11 and 12, or -12 and -11.

Example 4: A rectangle has an area of 96 cm². If the length is 4 cm more than the width, find the dimensions.
Let width =

x

, then length =

x + 4

Area:

x(x + 4) = 96 \Rightarrow x^2 + 4x - 96 = 0

Factor:

(x + 12)(x - 8) = 0 \Rightarrow x = 8

(reject -12)
Width = 8 cm, Length = 12 cm

Example 5: A man’s age is 4 years more than the square of his son’s age. If the sum of their ages is 36, find their ages.
Let son's age =

x

, man's age =

x^2 + 4

x + (x^2 + 4) = 36 \Rightarrow x^2 + x - 32 = 0

Factor:

(x + 8)(x - 4) = 0 \Rightarrow x = 4

Son = 4 years, Man = 20 years

Quadratic Equations in Rational and Geometric Settings

Example 6: A car travels a distance of 240 km in 2 hours. If its speed was reduced by 20 km/h, the journey would have taken 3 hours. Find the original speed.
Let original speed =

x

, then:

\frac{240}{x - 20} = 3 \Rightarrow \frac{240}{x} = 2

Cross-multiplying:

240(x - 20) = 3x \cdot 240 \Rightarrow x(x - 20) = 160

Form and solve the quadratic:

x^2 - 20x - 160 = 0

Practice Exercises

Solve $x^2 - 7x + 12 = 0$
Solve using the quadratic formula: $3x^2 + 5x - 2 = 0$
[WAEC] If the area of a square is the same as the area of a rectangle with length $x + 3$ and width $x - 2$ , find $x$ . [Past Question]
The product of two numbers is 120 and their difference is 2. Find the numbers.
A man’s age is 6 years more than twice his son's age. The product of their ages is 168. Find their ages.
A rectangular field is 3 m longer than it is wide. Its area is 88 m². Find its dimensions.
[NECO] A sum of money is shared between two people such that the square of one person’s share exceeds the other’s by ₦84. If the total is ₦120, find their shares. [Past Question]
The length of the hypotenuse of a right-angled triangle is 13 cm. One side is 5 cm longer than the other. Find the lengths of the two shorter sides.
[WAEC] The perimeter of a triangle is 30 cm. One side is twice the second and the third is 3 cm more than the second. Find the sides. [Past Question]
A garden's length is 4 m more than its width. If the area is 60 m², find its dimensions.

Conclusion/Recap

Advanced problem solving with quadratic equations sharpens your ability to translate real-world problems into algebraic models. These problems require a good understanding of the nature of quadratic expressions and their roots. Continuous practice will make you more confident in identifying and solving quadratic equations in diverse contexts.