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).
• Determine the nature of roots using the discriminant.
• Apply these skills to solve problems.

Lesson summary

This week, we delve deeper into the fascinating world of equations and inequalities. Specifically, we will focus on solving quadratic inequalities, simultaneous equations (linear and quadratic), and understanding the nature of roots of quadratic equations using the discriminant. These skills are crucial not just for mathematical understanding, but also for problem-solving in various real-life scenarios, from optimizing business profits to predicting population growth. Understanding how to manipulate equations and inequalities enables you to make informed decisions and analyze situations critically.

Lesson notes

2.1 Quadratic Inequalities A quadratic inequality is an inequality that can be written in the form ax² + bx + c > 0, ax² + bx + c 0 (or 0) or downwards (a

0. Standard form: The inequality is already in standard form.

Critical values: Solve x² - 3x - 4 =

0. Factoring gives (x - 4)(x + 1) = 0, so x = 4 or x = -

1. Sketch: a = 1 > 0, so the parabola opens upwards.

Intervals: We have three intervals to consider: x

4. We can test a value in each interval: x

0. So, x 4: Let x =

5. Then (5)² - 3(5) - 4 = 25 - 15 - 4 = 6 >

0. So, x > 4 is part of the solution.

Solution Set: The solution is x

4. In interval notation, this is (-∞, -1) ∪ (4, ∞).

Example 2: Solve the inequality -2x² + 8x - 6 ≤

0. Standard Form: Already in standard form.

Critical Values: Solve -2x² + 8x - 6 =

0. Divide by -2 to get x² - 4x + 3 =

0. Factoring gives (x - 1)(x - 3) = 0, so x = 1 or x =

3. Sketch: a = -2 3. x 0. x > 3: Let x =

4. Then -2(4)² + 8(4) - 6 = -32 + 32 - 6 = -6 ≤

0. Solution Set: Since the inequality is ≤ 0, we also include the roots. Thus, the solution is x ≤ 1 or x ≥

3. In interval notation: (-∞, 1] ∪ [3, ∞). 2.2 Simultaneous Equations (Linear and Quadratic) To solve simultaneous equations involving one linear and one quadratic equation, the primary method is substitution.

Steps: Solve the linear equation for one variable: Express one variable (e.g., y) in terms of the other (e.g., x).

Substitute: Substitute this expression into the quadratic equation.

Solve the resulting quadratic equation: This will give you the values of one variable (e.g., x).

Substitute back: Substitute the values you found in step 3 back into the linear equation to find the corresponding values of the other variable (e.g., y). Write the solutions as ordered pairs (x, y).

Example 3: Solve the simultaneous equations: y = x + 1 (linear) x² + y² = 5 (quadratic)

Solve the linear equation for y: y = x + 1 (already done)

Substitute: Substitute y = x + 1 into the quadratic equation: x² + (x + 1)² = 5 Solve the resulting quadratic: x² + x² + 2x + 1 = 5 2x² + 2x - 4 = 0 x² + x - 2 = 0 (x + 2)(x - 1) = 0 x = -2 or x = 1 Substitute back: If x = -2, then y = -2 + 1 = -

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

2. Solution Set: The solutions are (-2, -1) and (1, 2). 2.3 The Nature of Roots of a Quadratic Equation The discriminant (Δ) of a quadratic equation ax² + bx + c = 0 is given by: Δ = b² - 4ac The discriminant determines the nature of the roots: Δ > 0: Two distinct real roots. Δ = 0: Two equal real roots (one repeated root). *Δ 0, the equation has two distinct real roots.

Example 5: Determine the nature of the roots of the equation x² - 4x + 4 = 0. a = 1, b = -4, c = 4 Δ = (-4)² - 4(1)(4) = 16 - 16 = 0 Since Δ = 0, the equation has two equal real roots (one repeated root).

Example 6: Determine the nature of the roots of the equation x² + x + 1 = 0. a = 1, b = 1, c = 1 Δ = (1)² - 4(1)(1) = 1 - 4 = -3 Since Δ = -3 0).

Test the intervals: x 3. x 0. 2 3: Let x =

4. Then (4)² - 5(4) + 6 = 16 - 20 + 6 = 2 >

0. The solution is 2 =

0. Solution: Multiply both sides by -1: x^2 - 6x + 5 0).

Test the intervals: x 5. x 0. 1 5: Let x =

6. Then (6)^2 - 6(6) + 5 = 36 - 36 + 5 = 5 >

0. Since the inequality is

0. Solve the quadratic inequality -x² + 4x ≤

0. Solve the simultaneous equations: y = 3x + 2 and x² + y² =

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

4. Determine the nature of the roots of the quadratic equation x² + 6x + 9 =

0. Determine the nature of the roots of the quadratic equation 3x² - 2x + 1 =

0. Determine the values of k for which the equation x² + kx + 4 = 0 has two equal real roots.

Solve the inequality: (x-2)/(x+3) > 0 Solve for x: |2x - 1| 5, the farmer can determine the range of fertilizer amounts that will result in a profitable yield.

Designing a Bridge Arch: Civil engineers designing a bridge in the Drakensberg mountains need to ensure the arch's structural integrity. The shape of the arch can be represented by the equation y = -0.005x² + x, where y is the height of the arch at a distance x from one end. To ensure sufficient clearance for vehicles passing underneath, the height of the arch must be at least 10 meters. By solving the inequality -0.005x² + x ≥ 10, the engineers can determine the span of the arch that meets the clearance requirement.

Profit Maximization for a Small Business: A local craft vendor in Cape Town makes and sells beaded necklaces. The profit P (in Rand) depends on the number of necklaces x they sell, and can be modeled by the equation P = -0.2x² + 20x -

1

0

0. To make a sustainable income, the vendor needs to ensure the profit is greater than R

4

0

0. By solving the inequality -0.2x² + 20x - 100 > 400, they can determine the number of necklaces they need to sell each month to achieve their desired profit. Differentiation, Remediation and Extension Differentiation and Remediation: For struggling learners: Provide extra practice with simpler quadratic equations and inequalities (e.g., those that can be easily factored).