Lesson Notes By Weeks and Term v4 - SHS 1

Lesson Video

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Performance objectives

• Explain the process of completing the square.
• Derive the quadratic formula.
• Solve quadratic equations using the formula.
• Use the discriminant to find the nature of roots.
• Model real-life problems with quadratic equations.

Lesson summary

This lesson explores the powerful tools derived from quadratic equations, which are fundamental in various fields. We will move beyond basic factorisation to understand a universal method for solving any quadratic equation. We will discover how the famous quadratic formula is not magic but is derived logically through the method of completing the square. Understanding these applications is crucial for solving real-world problems, from calculating the path of a kicked football at the Accra Sports Stadium to helping a local entrepreneur in Kejetia Market maximize their profit.

Lesson notes

Part 1: The Heart of the Matter – Deriving the Quadratic Formula

We begin with the general form of any quadratic equation: $ax^2 + bx + c = 0$ (where $a \neq 0$)

Our goal is to rearrange this equation to solve for $x$. The method we will use is completing the square.

Step-by-Step Derivation: Divide by the leading coefficient, `a`: This makes the coefficient of $x^2$ equal to 1, which is necessary for completing the square. $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ Move the constant term to the right side: Isolate the terms containing $x$. $x^2 + \frac{b}{a}x = -\frac{c}{a}$ Complete the square on the left side: To make the left side a perfect square trinomial, we take half of the coefficient of $x$, square it, and add it to *both sides* of the equation. The coefficient of $x$ is $\frac{b}{a}$. Half of it is $\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}$. Squaring this gives $(\frac{b}{2a})^2 = \frac{b^2}{4a^2}$. Now, add this to both sides: $x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2$ Factor the left side and simplify the right side: The left side is now a perfect square. For the right side, find a common denominator. $(x + \frac{b}{2a})^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$ $(x + \frac{b}{2a})^2 = \frac{-4ac + b^2}{4a^2}$ $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$ Take the square root of both sides: Remember to include the positive and negative roots ($\pm$). $\sqrt{(x + \frac{b}{2a})^2} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$ $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$ $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ Isolate `x`: Subtract $\frac{b}{2a}$ from both sides to get the final formula. $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ Combine into a single fraction: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Evaluation guide

• Use the method of completing the square to transform any quadratic equation that has the same solutions and explain how the quadratic formula is derived from this form.
• Activity 1 : The quadratic formula and its usage.
• Collaborative Learning, Experiential Learning, Whole class discussion, and Problem -
• based Learning: