Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Solve quadratic inequalities in one variable.
• Solve simultaneous equations (one linear, one quadratic).
• Formulate and solve real-world word problems.
• Understand and apply the discriminant to determine the nature of roots.

Lesson summary

This week, we delve deeper into the world of equations and inequalities, building upon the foundational knowledge you gained in earlier grades. Equations and inequalities are fundamental tools in mathematics and are used extensively in various fields, from economics and engineering to everyday problem-solving. They allow us to model real-world situations, analyze relationships between quantities, and make informed decisions. For example, understanding inequalities helps a small business owner in Soweto determine the price range for their products to maximize profit, or allows a farmer in the Western Cape to calculate the optimal amount of fertilizer to use based on crop yield and cost.

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 and closed circles for ≤ and ≥.

Example 1: Solve x² - 3x - 4 >

0. The inequality is already in standard form.

Find the critical values: Solve x² - 3x - 4 =

0. Factoring gives: (x - 4)(x + 1) =

0. Therefore, x = 4 or x = -

1. Create a number line: Mark -1 and 4 on the number line.

Test intervals: Interval 1: x

0. So, this interval is part of the solution.

Interval 2: -1

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

0. So, this interval is part of the solution.

Solution set: x

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

Graphical Representation: A number line with an open circle at -1 and an open circle at 4, with arrows extending to the left from -1 and to the right from

4. Example 2: Solve 2x² + 5x ≤

3. Rewrite the inequality: 2x² + 5x - 3 ≤

0. Find the critical values: Solve 2x² + 5x - 3 =

0. Factoring gives: (2x - 1)(x + 3) =

0. Therefore, x = 1/2 or x = -

3. Create a number line: Mark -3 and 1/2 on the number line.

Test intervals: Interval 1: x

0. So, this interval is not part of the solution.

Interval 2: -3 1/

2. Let x = 1. 2(1)² + 5(1) - 3 = 2 + 5 - 3 = 4 >

0. So, this interval is not part of the solution.

Solution set: -3 ≤ x ≤ 1/

2. In interval notation: [-3, 1/2]. Notice that we included -3 and 1/2 because of the "≤" sign.

Graphical Representation: A number line with a closed circle at -3 and a closed circle at 1/2, with a line segment connecting them. 2.2 Simultaneous Equations (One Linear, One Quadratic) These are systems of equations where one equation is linear (degree 1) and the other is quadratic (degree 2).

Solving Simultaneous Equations: The method involves substitution.

Steps: Solve the linear equation for one variable: Express one variable (usually y or x) in terms of the other.

Substitute into the quadratic equation: Substitute the expression obtained in step 1 into the quadratic equation. This will result in a quadratic equation in one variable.

Solve the quadratic equation: Solve the resulting quadratic equation using factoring, the quadratic formula, or completing the square. Find the corresponding values of the other variable: Substitute the solutions obtained in step 3 back into the linear equation to find the corresponding values of the other variable.

Write the solutions as ordered pairs: Express the solutions as ordered pairs (x, y).

Example 3: Solve the following system of equations: y = x + 1 (Linear) x² + y² = 13 (Quadratic) The linear equation is already solved for y. Substitute y = x + 1 into the quadratic equation: x² + (x + 1)² =

1

3. Solve the quadratic equation: x² + x² + 2x + 1 = 13 2x² + 2x - 12 = 0 x² + x - 6 = 0 (Divide by 2) (x + 3)(x - 2) = 0 Therefore, x = -3 or x =

2. Find the corresponding values of y: If x = -3, then y = -3 + 1 = -

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

3. Solution set: The solutions are (-3, -2) and (2, 3). 2.3 The Discriminant The discriminant, denoted by Δ (Delta), is the part of the quadratic formula under the square root: Δ = b² - 4ac, where a, b, and c are the coefficients of the quadratic equation ax² + bx + c =

0. The discriminant tells us about the nature of the roots of the quadratic equation. Δ > 0: The equation has two distinct real roots. Δ = 0: The equation has one real root (a repeated root). *Δ 0. 2 3: Let x = 4. (4)² - 5(4) + 6 = 16 - 20 + 6 = 2 >

0. Solution set: 2 x² + 2x - 4 =

0. Using the quadratic formula: x = (-2 ± √(2² - 4(1)(-4))) / (2(1)) = (-2 ± √(20)) / 2 = (-2 ± 2√5) / 2 = -1 ± √

5. Find the corresponding values of y: If x = -1 + √5, then y = 2(-1 + √5) - 1 = -2 + 2√5 - 1 = -3 + 2√

5. If x = -1 - √5, then y = 2(-1 - √5) - 1 = -2 - 2√5 - 1 = -3 - 2√

5. Solution set: (-1 + √5, -3 + 2√5) and (-1 - √5, -3 - 2√5).

Commentary: Because the quadratic didn't factor neatly, we had to use the quadratic formula. Remember to substitute each solution back into the linear equation to find the matching y-value.

Question 3: Determine the nature of the roots of the equation 3x² - 2x + 1 =

0. Solution: a = 3, b = -2, c = 1 Δ = (-2)² - 4(3)(1) = 4 - 12 = -

8. Since Δ 0