Scheme of work: IGCSE Higher: Year 10: Term 5: Quadratic Equations

#### Prerequisite Knowledge

1. Solving Linear Equations
• Concept: Understanding how to solve linear equations where the unknown appears on one side or both sides of the equation.
• Example: $\text{Solve } 2x + 3 = 7.$ $2x = 4 \rightarrow x = 2.$ $\text{Solve } 3x – 5 = x + 1.$ $3x – x = 1 + 5 \rightarrow 2x = 6 \rightarrow x = 3.$
2. Expanding and Factorising Quadratic Expressions
• Concept: Knowing how to expand and factorise quadratic expressions.
• Example (Expanding): $\text{Expand } (x + 2)(x – 3).$ $x^2 – 3x + 2x – 6 = x^2 – x – 6.$
• Example (Factorising): $\text{Factorise } x^2 – 5x + 6.$ $x^2 – 5x + 6 = (x – 2)(x – 3).$
3. Changing the Subject of a Formula
• Concept: Understanding how to rearrange equations to make a different variable the subject.
• Example: $\text{Rearrange } y = 3x + 4 \text{ to make } x \text{ the subject.}$ $y – 4 = 3x \rightarrow x = \frac{y – 4}{3}.$

#### Success Criteria

1. Solve Quadratic Equations by Factorisation
• Objective: Accurately solve quadratic equations by expressing them in the form $$(x – p)(x – q) = 0$$ and finding the values of $$x$$ that satisfy the equation.
• Example: $\text{Solve } x^2 – 5x + 6 = 0 \text{ by factorisation.}$ $x^2 – 5x + 6 = (x – 2)(x – 3) = 0.$ $\text{So, } x – 2 = 0 \text{ or } x – 3 = 0 \rightarrow x = 2 \text{ or } x = 3.$
2. Solve Quadratic Equations by Using the Quadratic Formula or Completing the Square
• Objective: Accurately solve quadratic equations using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$ or by completing the square.
• Example (Quadratic Formula): $\text{Solve } x^2 – 4x – 5 = 0 \text{ using the quadratic formula.}$ $a = 1, b = -4, c = -5.$ $x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}.$ $x = 5 \text{ or } x = -1.$
• Example (Completing the Square): $\text{Solve } x^2 – 6x + 5 = 0 \text{ by completing the square.}$ $x^2 – 6x + 5 = (x – 3)^2 – 4 = 0.$ $(x – 3)^2 = 4 \rightarrow x – 3 = \pm 2 \rightarrow x = 5 \text{ or } x = 1.$
3. Form and Solve Quadratic Equations from Data Given in a Context
• Objective: Formulate quadratic equations from word problems or real-life situations and solve them using appropriate methods.
• Example: $\text{A rectangle has an area of 30 square units and its length is 5 units more than its width. Find the dimensions of the rectangle.}$ $\text{Let the width be } x. \text{ Then the length is } x + 5.$ $x(x + 5) = 30 \rightarrow x^2 + 5x – 30 = 0.$ $\text{Solve } x^2 + 5x – 30 = 0 \text{ by the quadratic formula:}$ $x = \frac{-5 \pm \sqrt{5^2 + 4 \cdot 30}}{2} = \frac{-5 \pm \sqrt{145}}{2}.$
4. Solve Simultaneous Equations in Two Unknowns, One Equation Being Linear and the Other Being Quadratic
• Objective: Solve simultaneous equations where one equation is linear and the other is quadratic by substitution or elimination methods.
• Example: $\text{Solve the system:}$ $y = 2x + 3$ $y = x^2 + x + 1.$ $\text{Substitute } y = 2x + 3 \text{ into } y = x^2 + x + 1.$ $2x + 3 = x^2 + x + 1 \rightarrow x^2 – x – 2 = 0.$ $(x – 2)(x + 1) = 0 \rightarrow x = 2 \text{ or } x = -1.$ $\text{For } x = 2, y = 2(2) + 3 = 7.$ $\text{For } x = -1, y = 2(-1) + 3 = 1.$ $\text{Solutions: } (2, 7) \text{ and } (-1, 1).$

#### Key Concepts

1. Solve Quadratic Equations by Factorisation
• Concept: Quadratic equations can often be solved by factorising them into two binomial expressions and setting each factor to zero.
2. Solve Quadratic Equations by Using the Quadratic Formula or Completing the Square
• Concept: The quadratic formula $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$ can solve any quadratic equation. Completing the square is another method that can make it easier to solve or graph a quadratic equation.
3. Form and Solve Quadratic Equations from Data Given in a Context
• Concept: Many real-world problems can be modeled by quadratic equations. Forming these equations requires translating a word problem into an algebraic expression.
4. Solve Simultaneous Equations in Two Unknowns, One Equation Being Linear and the Other Being Quadratic
• Concept: Solving simultaneous equations where one is linear and the other is quadratic often involves substituting the linear equation into the quadratic equation to reduce it to a single-variable equation.

#### Common Misconceptions

1. Solve Quadratic Equations by Factorisation
• Common Mistake: Incorrectly factorising the quadratic equation or missing a factor.
• Example: $\text{Incorrect: For } x^2 – 5x + 6 = 0, \text{ factorising as } (x – 1)(x – 6) = 0.$ $\text{Correct: Factorise correctly as } (x – 2)(x – 3) = 0 \rightarrow x = 2 \text{ or } x = 3.$
2. Solve Quadratic Equations by Using the Quadratic Formula or Completing the Square
• Common Mistake: Misapplying the quadratic formula or making arithmetic errors during calculations.
• Example (Quadratic Formula): $\text{Incorrect: For } x^2 – 4x – 5 = 0, \text{ using incorrect values: } x = \frac{-4 \pm \sqrt{16 – 20}}{2}.$ $\text{Correct: Use correct values: } x = \frac{-(-4) \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}.$
• Example (Completing the Square): $\text{Incorrect: For } x^2 – 6x + 5 = 0, \text{ completing the square incorrectly: } (x – 3)^2 + 4 = 0.$ $\text{Correct: Complete the square correctly: } x^2 – 6x + 5 = (x – 3)^2 – 4 = 0.$ $(x – 3)^2 = 4 \rightarrow x – 3 = \pm 2 \rightarrow x = 5 \text{ or } x = 1.$
3. Form and Solve Quadratic Equations from Data Given in a Context
• Common Mistake: Misinterpreting the problem or setting up the wrong equation.
• Example: $\text{Incorrect: For a rectangle problem, interpreting the length as } x – 5 \text{ instead of } x + 5.$ $\text{Correct: Set up the equation based on correct interpretation: }$ $\text{Let width be } x, \text{ then length is } x + 5.$ $x(x + 5) = 30 \rightarrow x^2 + 5x – 30 = 0.$ $\text{Solve correctly: } x = \frac{-5 \pm \sqrt{5^2 + 4 \cdot 30}}{2} = \frac{-5 \pm \sqrt{145}}{2}.$
4. Solve Simultaneous Equations in Two Unknowns, One Equation Being Linear and the Other Being Quadratic
• Common Mistake: Making errors in substitution or simplification, leading to incorrect solutions.
• Example: $\text{Incorrect: Given } y = 2x + 3 \text{ and } y = x^2 + x + 1, \text{ substituting incorrectly: } 2x + 3 = x^2 – x + 1.$ $\text{Correct: Substitute correctly: } 2x + 3 = x^2 + x + 1.$ $\text{Solve correctly: } x^2 – x – 2 = 0 \rightarrow (x – 2)(x + 1) = 0.$ $\text{Find corresponding } y \text{ values: For } x = 2, y = 2(2) + 3 = 7.$ $\text{For } x = -1, y = 2(-1) + 3 = 1.$ $\text{Solutions: } (2, 7) \text{ and } (-1, 1).$

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