Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Solve quadratic inequalities algebraically and graphically.
• Solve simultaneous equations (one linear, one quadratic).
• Formulate and solve equations from word problems.
• Use the discriminant to determine the nature of roots.
• Solve word problems involving quadratic equations.

Lesson summary

This week, we delve deeper into equations and inequalities, building upon your Grade 10 foundation and expanding your problem-solving skills. Equations and inequalities are fundamental tools in mathematics and are crucial for modelling real-world scenarios, especially in fields like finance, science, and engineering. Being able to solve equations and inequalities allows us to determine unknown quantities, optimize resources, and make informed decisions. In the South African context, these skills are invaluable for understanding economic models, managing personal finances, and analyzing data related to social issues.

Lesson notes

Quadratic Inequalities A quadratic inequality is an inequality that can be written in the general form: `ax² + bx + c > 0` `ax² + bx + c 0`, the parabola opens upwards (minimum turning point). If `a or 0`. The inequality is already in the required form. Solve `x² - 3x - 4 = 0`. Factoring gives `(x - 4)(x + 1) = 0`.

Therefore, `x = 4` or `x = -1`.

Sketch a graph: Since `a = 1 > 0`, the parabola opens upwards. The x-intercepts are -1 and

4. The inequality `x² - 3x - 4 > 0` asks for values of x where the parabola is above the x-axis (positive values). This occurs when `x 4`.

Solution: `x 4`, or in interval notation, `(-∞, -1) ∪ (4, ∞)`.

Example 2: Solve `-2x² + 6x ≤ 0`. The inequality is in the required form. Solve `-2x² + 6x = 0`. Factoring gives `-2x(x - 3) = 0`.

Therefore, `x = 0` or `x = 3`.

Sketch a graph: Since `a = -2 `2x² + 2x - 12 = 0` => `x² + x - 6 = 0` Factor: `(x + 3)(x - 2) = 0`.

Therefore, `x = -3` or `x = 2`. Substitute `x = -3` into `y = x + 1`: `y = -3 + 1 = -2`. Substitute `x = 2` into `y = x + 1`: `y = 2 + 1 = 3`.

Solution: `(-3, -2)` and `(2, 3)`. The Discriminant and Nature of Roots The discriminant, denoted by Δ (delta), is a part of the quadratic formula that helps determine the nature of the roots of a quadratic equation `ax² + bx + c = 0`.

The discriminant is given by: `Δ = b² - 4ac` The discriminant tells us the following about the roots: Δ > 0: The quadratic equation has two distinct real roots. The parabola intersects the x-axis at two different points. Δ = 0: The quadratic equation has one real (repeated) root (also sometimes described as two equal real roots). The parabola touches the x-axis at one point (the vertex). Δ 0 and a perfect square: The quadratic equation has two distinct rational real roots. The roots are real and can be written as a ratio of two integers. Δ > 0 and NOT a perfect square: The quadratic equation has two distinct irrational real roots. The roots are real but cannot be written as a ratio of two integers.

Example 4: Determine the nature of the roots of `2x² - 5x + 3 = 0`. `Δ = b² - 4ac = (-5)² - 4(2)(3) = 25 - 24 = 1` Since `Δ = 1 > 0` and is a perfect square, the equation has two distinct, rational, real roots. Word Problems Involving Quadratic Equations Solving word problems involving quadratic equations involves translating the problem's description into a mathematical equation.

Read the problem carefully: Understand what the problem is asking.

Define variables: Assign variables to the unknown quantities.

Formulate the equation: Translate the word problem into a quadratic equation using the defined variables.

Solve the equation: Use factoring, completing the square, or the quadratic formula to solve the equation.

Interpret the solutions: Check if the solutions are reasonable in the context of the problem. Discard any solutions that don't make sense (e.g., negative lengths).

Answer the question: State the answer in a clear and concise sentence, including the appropriate units.

Example 5: A rectangular garden has a length that is 3 meters longer than its width. If the area of the garden is 10 square meters, find the width and length of the garden. Let `w` be the width of the garden in meters. The length is `w + 3` meters. The area is given by `length × width`, so `w(w + 3) = 10`.

Expand and simplify: `w² + 3w = 10` => `w² + 3w - 10 = 0`.

Factor: `(w + 5)(w - 2) = 0`.

Therefore, `w = -5` or `w = 2`. Since the width cannot be negative, `w = 2` meters. The length is `w + 3 = 2 + 3 = 5` meters.

Answer: The width of the garden is 2 meters, and the length is 5 meters. Guided Practice (With Solutions)

Question 1: Solve the inequality `x² + 2x - 8 ≤ 0`.

Solution: Find the critical values: `x² + 2x - 8 = 0` => `(x + 4)(x - 2) = 0` => `x = -4` or `x = 2`.

Sketch the graph: Parabola opens upwards (a > 0). x-intercepts are -4 and

2. Determine the intervals: The inequality `x² + 2x - 8 ≤ 0` asks for where the parabola is below or on the x-axis. This is between -4 and 2, inclusive.

Solution: `-4 ≤ x ≤ 2`, or `[-4, 2]`.

Question 2: Solve the simultaneous equations: `y = 2x - 1` and `x² + y = 4`.

Solution: Substitute `y = 2x - 1` into `x² + y = 4`: `x² + (2x - 1) = 4` Simplify: `x² + 2x - 1 = 4` => `x² + 2x - 5 = 0`.

Use the quadratic formula to solve for x: `x = (-b ± √(b² - 4ac)) / (2a) = (-2 ± √(2² - 4(1)(-5))) / (2(1)) = (-2 ± √(24)) / 2 = (-2 ± 2√6) / 2 = -1 ± √6` So `x = -1 + √6` or `x = -1 - √6`. Now substitute these back into `y = 2x - 1`: If `x = -1 + √6`, `y = 2(-1 + √6) - 1 = -2 + 2√6 - 1 = -3 + 2√6`. If `x = -1 - √6`, `y = 2(-1 - √6) - 1 = -2 - 2√6 - 1 = -3 - 2√6`.

Solution: `(-1 + √6, -3 + 2√6)` and `(-1 - √6, -3 - 2√6)`.

Question 3: Determine the nature of the roots of `x² - 6x + 9 = 0`.

Solution: `Δ = b² - 4ac = (-6)² - 4(1)(9) = 36 - 36 = 0`. Since `Δ = 0`, the equation has one real (repeated) root.

Question 4: A farmer wants to fence a rectangular area for his goats. He has 40 meters of fencing available. What dimensions will maximize the area enclosed?