Quadratic Equations

Grade 12 Math - Quadratic Equations

Lesson Objectives

Solve quadratic equations using advanced algebraic techniques.
Apply methods like completing the square and the quadratic formula.
Solve simultaneous equations involving linear and quadratic expressions.
Analyze real-world and abstract problems using quadratic models.
Interpret the nature of roots using the discriminant.

Lesson Introduction

Quadratic equations arise in various real-life problems — from physics and engineering to business and design. Understanding how to manipulate and solve them is essential. In this lesson, we will explore techniques for solving quadratics and systems involving quadratic and linear equations.

Core Lesson Content

Worked Example

Example 1: Solve

x^2 - 7x + 12 = 0

.
Factor the quadratic:

x^2 - 7x + 12 = (x - 3)(x - 4) = 0

Therefore,

x = 3 \text{ or } x = 4

Example 2: Solve

2x^2 + 5x - 3 = 0

using the quadratic formula.

a = 2, b = 5, c = -3

x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}

x = \frac{2}{4} = 0.5, \quad x = \frac{-12}{4} = -3

Example 3: Solve

x^2 - 6x = 7

by completing the square.
Bring to standard form:

x^2 - 6x - 7 = 0

Complete the square:

x^2 - 6x + 9 = 7 + 9 = 16

(x - 3)^2 = 16 \Rightarrow x - 3 = \pm4 \Rightarrow x = 7 \text{ or } x = -1

Example 4: Find the roots of

3x^2 + x - 10 = 0

.
Using the quadratic formula:

x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-10)}}{2(3)} = \frac{-1 \pm \sqrt{121}}{6}

x = \frac{-1 \pm 11}{6} \Rightarrow x = \frac{10}{6} = \frac{5}{3}, \quad x = \frac{-12}{6} = -2

Example 5: Solve simultaneously:

y = 2x + 1

y = x^2 + x - 6

.
Substitute

y

2x + 1 = x^2 + x - 6 \Rightarrow x^2 - x - 7 = 0

x = \frac{1 \pm \sqrt{1 + 28}}{2} = \frac{1 \pm \sqrt{29}}{2}

Get approximate roots and plug into

y = 2x + 1

Example 6: Find

x

2x^2 = 18x

.
Rearranging:

2x^2 - 18x = 0 \Rightarrow 2x(x - 9) = 0

x = 0 \text{ or } x = 9

Example 7: Solve

(x + 2)^2 = 25

x + 2 = \pm5 \Rightarrow x = 3 \text{ or } -7

Example 8: Solve

x^2 = -4

.
No real solution since square of a real number cannot be negative. Hence, solution is imaginary.

Example 9: If

x^2 + px + 16 = 0

has equal roots, find

p

.
Discriminant

D = p^2 - 4(1)(16) = 0 \Rightarrow p^2 = 64 \Rightarrow p = \pm8

Example 10: Solve simultaneously:

x^2 + y^2 = 25

y = x + 1

Substitute:

x^2 + (x + 1)^2 = 25

x^2 + x^2 + 2x + 1 = 25 \Rightarrow 2x^2 + 2x - 24 = 0 \Rightarrow x^2 + x - 12 = 0

x = 3 \text{ or } -4 \Rightarrow y = 4 \text{ or } -3

Exercises

Solve $x^2 + 4x + 3 = 0$ .
$3x^2 - x - 4 = 0$ , solve by completing the square.
Solve $x^2 - 10x + 21 = 0$ using factorization.
[WAEC] Solve: $x^2 + x - 6 = 0$ (Past Question)
[NECO] Solve: $x^2 = 5x + 6$ (Past Question)
Solve simultaneously: $y = x^2 - 3$ , $y = 2x + 1$
Solve: $x(x + 5) = 14$
Solve: $5x^2 + 20x + 15 = 0$
[JAMB] If $x^2 + kx + 9 = 0$ has one real root, find $k$ . (Past Question)
[WAEC] Solve simultaneously: $y = 3x + 2, y = x^2 + x - 4$ (Past Question)

Conclusion/Recap

In this lesson, we explored advanced ways to solve quadratic equations and systems involving them. These methods — factoring, completing the square, the quadratic formula, and solving simultaneous equations — are essential tools in further mathematics.