Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Solve quadratic inequalities algebraically and graphically.
• Solve inequalities involving rational expressions.
• Represent solutions using interval notation and on a number line.
• Apply inequalities to solve problems.

Lesson summary

This week, we delve deeper into equations and inequalities, building on your prior knowledge. We will focus specifically on solving quadratic inequalities and inequalities with rational expressions. Understanding these concepts is crucial not only for further mathematics studies but also for modeling real-world scenarios like optimizing profits in a small business or determining safe operating ranges for machinery – situations relevant to many South African contexts. Many fields, from agriculture (optimizing crop yields) to engineering (designing safe structures) rely heavily on the ability to manipulate and solve equations and inequalities.

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 or

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

Critical values: Solve x² - 3x - 4 =

0. Factoring, we get (x - 4)(x + 1) =

0. So, x = 4 and x = -1 are the critical values.

Sign table/Number line: | Interval | Test Value | x² - 3x - 4 | Sign | | :------- | :--------- | :---------- | :--- | | x 4 | x = 5 | (5)² - 3(5) - 4 = 6 | + | Alternatively, visualize a parabola. Since the coefficient of x² is positive, it opens upwards. It crosses the x-axis at x=-1 and x=

4. It is above the x-axis (positive) when x 4, and below (negative) between -1 and

4. Solution set: We want x² - 3x - 4 > 0, so we look for the intervals where the expression is positive.

Interval Notation: The solution is x 4, which in interval notation is (-∞, -1) ∪ (4, ∞).

Example 2: Solve the inequality -2x² + 5x + 3 ≤

0. Standard Form: The inequality is in standard form. It's often helpful (but not necessary) to multiply by -1 to make the x² term positive: 2x² - 5x - 3 ≥

0. Remember to flip the inequality sign when multiplying by a negative number.

Critical Values: Solve 2x² - 5x - 3 =

0. Factoring, we get (2x + 1)(x - 3) =

0. So x = -1/2 and x = 3 are critical values.

Sign Table/Number Line: | Interval | Test Value | 2x² - 5x - 3 | Sign | | :------- | :--------- | :---------- | :--- | | x 3 | x = 4 | 2(4)² - 5(4) - 3 = 9 | + | Solution Set: We want 2x² - 5x - 3 ≥ 0, so we're looking for where it's positive or zero.

Interval Notation: x ≤ -1/2 or x ≥ 3, which is (-∞, -1/2] ∪ [3, ∞). Note the square brackets because the inequality includes equality. Inequalities with Rational Expressions A rational inequality is an inequality involving rational expressions (fractions with polynomials in the numerator and denominator).

Solving Rational Inequalities: Rewrite the inequality: Manipulate the inequality so that one side is zero and the other side is a single rational expression. This might involve finding a common denominator.

Find the critical values: Determine the values of x that make the numerator equal to zero and the values of x that make the denominator equal to zero. These are your critical values.

Create a sign table or use a number line: Draw a number line and mark the critical values on it. These divide the number line into intervals. Choose a test value from each interval. Substitute each test value into the simplified rational expression. Determine if the expression is positive or negative for each test value.

Identify the solution set: Based on the sign table or number line, identify the intervals where the inequality is satisfied.

Remember: If the inequality is strict (> or

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

Critical values: Numerator: x + 2 = 0 => x = -

2. Denominator: x - 3 = 0 => x =

3. Sign table/Number line: | Interval | Test Value | (x + 2) / (x - 3) | Sign | | :------- | :--------- | :----------------- | :--- | | x 3 | x = 4 | (4 + 2) / (4 - 3) = 6 | + | Solution set: We want (x + 2) / (x - 3) > 0, so we look for the intervals where the expression is positive.

Interval Notation: The solution is x 3, which in interval notation is (-∞, -2) ∪ (3, ∞). Notice the parentheses on both values because the inequality is strict and x=3 makes the denominator zero.

Example 4: Solve the inequality (x - 1) / (x + 2) ≤

2. Rewrite the inequality: (x - 1) / (x + 2) - 2 ≤

0. Find a common denominator: (x - 1 - 2(x + 2)) / (x + 2) ≤

0. Simplify: (x - 1 - 2x - 4) / (x + 2) ≤ 0 => (-x - 5) / (x + 2) ≤

0. We can also multiply both sides by -1 to have a positive x. (x+5)/(x+2) >= 0 Critical values: Numerator: -x - 5 = 0 => x = -

5. Denominator: x + 2 = 0 => x = -

2. Sign table/Number line: | Interval | Test Value | (x + 5) / (x + 2) | Sign | | :------- | :--------- | :----------------- | :--- | | x -2 | x = 0 | (0 + 5) / (0 + 2) = 5/2 | + | Solution set: We want (x + 5)/(x + 2) >= 0, so we look for where it's positive or zero.

Interval Notation: The solution is x ≤ -5 or x > -2, which in interval notation is (-∞, -5] ∪ (-2, ∞). Note the square bracket on -5 as the inequality includes the equality. Guided Practice (With Solutions)

Question 1: Solve the quadratic inequality x² + 2x - 8 2 | x = 3 | (3)² + 2(3) - 8 = 7 | + | Solution set: We want x² + 2x - 8 x = 1/

2. Denominator: x + 3 = 0 => x = -

3. Sign table/Number line: | Interval | Test Value | (2x - 1) / (x + 3) | Sign | | :------- | :--------- | :----------------- | :--- | | x 1/2 | x = 1 | (2(1) - 1) / (1 + 3) = 1/4 | + | Solution set: We want (2x - 1) / (x + 3) ≥ 0, so we look for the intervals where the expression is positive or zero.

Interval Notation: The solution is x 5 | x = 6 | (6)² - 6(6) + 5 = 5 | + | Solution set: We want x² - 6x + 5 ≤ 0, so we look for the intervals where the expression is negative or zero.

Interval Notation: The solution is 1 ≤ x ≤ 5, which in interval notation is [1, 5].