Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Solve quadratic equations by factorisation and formula.
• Solve quadratic inequalities and represent solutions.
• Solve simultaneous equations (one linear, one quadratic).
• Apply knowledge to solve real-world problems.
• Identify the nature of roots using the discriminant.

Lesson summary

This week, we delve deeper into the world of equations and inequalities. This is a crucial topic in mathematics as it provides us with the tools to model and solve real-world problems involving relationships between quantities. Understanding equations and inequalities allows us to make informed decisions in various aspects of life, from budgeting and financial planning to understanding scientific data and engineering challenges.

Think about it: calculating the cheapest cell phone data bundle given your usage, determining if you have enough money to buy groceries, or understanding if a bridge design can handle a certain weight load – all rely on the principles we will be learning.

Lesson notes

2.1 Quadratic Equations A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants and a ≠

0. The solutions to a quadratic equation are called its roots. We will focus on solving these equations using factorisation and the quadratic formula. 2.1.1 Solving by Factorisation Factorisation involves expressing the quadratic expression as a product of two linear factors. The principle used is that if the product of two numbers is zero, then at least one of the numbers must be zero.

Example 1: Solve x² + 5x + 6 = 0 Factorise: Find two numbers that multiply to 6 and add to

5. These are 2 and

3. So, the equation becomes (x + 2)(x + 3) =

0. Set each factor to zero: x + 2 = 0 or x + 3 =

0. Solve for x: x = -2 or x = -

3. Therefore, the roots are -2 and -

3. Example 2: Solve 2x² - 7x + 3 = 0 Factorise: This one requires a bit more work. We need two numbers that multiply to (2)(3) = 6 and add to -

7. These are -1 and -

6. Rewrite the middle term: 2x² - x - 6x + 3 =

0. Factor by grouping: x(2x - 1) - 3(2x - 1) =

0. This gives (2x - 1)(x - 3) =

0. Set each factor to zero: 2x - 1 = 0 or x - 3 =

0. Solve for x: x = 1/2 or x =

3. Therefore, the roots are 1/2 and 3. 2.1.2 Solving using the Quadratic Formula When factorisation is difficult or impossible, we use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a Example 3: Solve x² - 4x + 2 = 0 Identify a, b, and c: a = 1, b = -4, c =

2. Substitute into the formula: x = (4 ± √((-4)² - 4 1 2)) / (2 * 1)

Simplify: x = (4 ± √(16 - 8)) / 2 = (4 ± √8) / 2 = (4 ± 2√2) / 2 = 2 ± √2 The solutions are: x = 2 + √2 or x = 2 - √2. (Approximately, x = 3.41 or x = 0.59) 2.1.3 The Discriminant (b² - 4ac) The discriminant, Δ = b² - 4ac, tells us about the nature of the roots of the quadratic equation: If Δ > 0: The equation has two distinct real roots. If Δ = 0: The equation has one real root (a repeated root or two equal roots). If Δ 0 (or 0 Factorise (or use the quadratic formula to find roots): (x - 4)(x + 1) >

0. The roots are x = 4 and x = -

1. Sketch a parabola: The parabola opens upwards (since the coefficient of x² is positive). Mark the roots -1 and 4 on the x-axis.

Determine the intervals: We want the values of x where the parabola is above the x-axis (greater than 0). This occurs when x

4. Interval notation: The solution is (-∞, -1) ∪ (4, ∞).

Example 6: Solve -x² + 2x + 3 ≤ 0 Multiply by -1 (and reverse the inequality): x² - 2x - 3 ≥

0. Factorise: (x - 3)(x + 1) ≥

0. The roots are x = 3 and x = -

1. Sketch a parabola: The parabola opens upwards. Mark the roots -1 and 3 on the x-axis.

Determine the intervals: We want the values of x where the parabola is above or on the x-axis (greater than or equal to 0). This occurs when x ≤ -1 or x ≥

3. Interval notation: The solution is (-∞, -1] ∪ [3, ∞). Notice the square brackets indicating inclusion of the endpoints. 2.3 Simultaneous Equations (One Linear, One Quadratic) To solve simultaneous equations where one equation is linear and the other is quadratic: Solve the linear equation for one variable (e.g., solve for y in terms of x). Substitute this expression into the quadratic equation. Solve the resulting quadratic equation for the remaining variable (e.g., solve for x). Substitute the values of x back into the linear equation to find the corresponding values of y. Write the solutions as ordered pairs (x, y).

Example 7: Solve the following system: y = x + 1 x² + y² = 13 Substitute: Substitute y = x + 1 into the second equation: x² + (x + 1)² = 13 Simplify and solve for x: x² + x² + 2x + 1 = 13 => 2x² + 2x - 12 = 0 => x² + x - 6 = 0 => (x + 3)(x - 2) =

0. So, x = -3 or x =

2. Solve for y: If x = -3, y = -3 + 1 = -

2. If x = 2, y = 2 + 1 =

3. The solutions are: (-3, -2) and (2, 3). Guided Practice (With Solutions)

Question 1: Solve the equation 3x² - 5x - 2 = 0 by factorisation.

Solution: We need to find two numbers that multiply to (3)(-2) = -6 and add to -

5. These are -6 and

1. Rewrite the middle term: 3x² - 6x + x - 2 = 0 Factor by grouping: 3x(x - 2) + 1(x - 2) = 0 (3x + 1)(x - 2) = 0 3x + 1 = 0 or x - 2 = 0 x = -1/3 or x =

2. Commentary: This question tests basic factorisation skills. The key is to correctly identify the two numbers needed to split the middle term.

Question 2: Solve the equation x² + 3x - 5 = 0 using the quadratic formula. Give your answer to two decimal places.

Solution: a = 1, b = 3, c = -5 x = (-3 ± √(3² - 4 1 -5)) / (2 * 1) x = (-3 ± √(9 + 20)) / 2 x = (-3 ± √29) / 2 x = (-3 + √29) / 2 ≈ 1.19 or x = (-3 - √29) / 2 ≈ -4.19

Commentary: This question tests the ability to correctly apply the quadratic formula. Ensure you substitute the values accurately and simplify the expression correctly. Rounding to two decimal places is specified in the question, so pay attention to that.