Lesson Objectives

Dear learner; At the end of the lesson you will be able to:

• Identify quadratic inequalities and rewrite them in standard forms.
• Factor quadratic expressions and analyze the signs of the factors.
• Apply product properties to solve quadratic inequalities.
• Express solutions of quadratic inequalities in interval notation.

Key Terms

• quadratic inequality
• Property of products
• soluttion techniques
• product rules
• sign chart method

Brainstorming Questions

1. How can quadratic inequalities be used to optimize the dimensions of a parabolic reflector to maximize signal strength while maintaining structural integrity?

Solution
In engineering, especially in the design of parabolic reflectors (e.g., satellite dishes), the shape of the reflector is crucial for focusing signals effectively. The reflector can be modeled by a quadratic function, and its dimensions (e.g., width and height) must satisfy certain constraints to ensure it focuses signals optimally. For instance, the quadratic inequality might represent the height ( h ) of the reflector as a function of its width ( w ), ensuring that the structural strength ( S ) is maintained.

Application: Given the quadratic inequality $ax^2 + bx + c \leq 0$, where ( x ) represents one dimension of the reflector and $ax^2 + bx + c $ represents the strength function that must be non-positive for structural integrity, engineers can solve the inequality to find acceptable dimensions that maximize signal focus while ensuring the reflector’s strength is within safe limits.

2. How can quadratic inequalities help in determining the feasible budget range for a project where costs are modeled by a quadratic function?

Solution:
In project planning, the total cost ( C ) of a project may be represented by a quadratic function due to varying expenses like material costs and labor. For example, if the budget constraint is modeled by $ ax^2 + bx + c \leq B$, where ( B ) is the maximum allowable budget, solving this quadratic inequality helps determine the range of project variables (e.g., quantity of materials) that fit within the budget.

Application: By solving the quadratic inequality $ ax^2 + bx + c \leq B$, project managers can identify the feasible range for project parameters (like quantity) that stay within budget. For instance, if $2x^2 + 5x – 3 \leq 100$, solving this inequality determines the range of ( x ) that keeps the total cost below 100 US Dollar.

4.1. Quadratic Inequalities

Dear learner; Quadratic inequalities are mathematical expressions that involve quadratic functions and inequalities. These inequalities typically take the form of $ax^2 + bx + c < 0, ax^2 + bx + c > 0, ax^2 + bx + c ≤ 0, or ax^2 + bx + c ≥ 0$, where a, b, and c are constants and x is a variable. The solutions to quadratic inequalities are the values of x that satisfy the inequality. Graphically, these solutions can be represented on a number line or a coordinate plane. Quadratic inequalities are commonly used in algebra and calculus to analyze and solve real-world problems involving quadratic functions.

4.1.1. Solving quadratic inequalities using product properties

Product properties:

1. Rewrite the quadratic inequality in the form $ax^2 + bx + c > 0$ in the form of $mn > 𝟎$. Then $m.n > 0$, if and only if

                         i. m > 𝟎 and n > 𝟎 or                ii. m < 𝟎 and n < 𝟎

2. Rewrite the quadratic inequality in the form $ax^2 + bx + c < 0$ in the form of m ∙ n < 𝟎. Then, m ∙ n < 𝟎 if and only if

                  i. m > 𝟎 and n < 𝟎 or                ii. m < 𝟎 and n > 0

Example 1:– Solve $(x + 4)(x – 3) \leqslant 0$

Solution: According to product property 2, to be

m ∙ n < 𝟎 if and only if i. m > 𝟎 and n < 𝟎

case i. $x + 4 \leqslant 0$ and $x – 3 \geqslant 0$

$x \leqslant – 4$ and $x \geqslant 3$ . Thus, there are no any intersection between this two intervals. Answer: no solution.

or case ii. $x + 4 \geqslant 0$ and $x – 3 \leqslant 0$

x \geqslant – 4 and x \leqslant 3 . Thus, the intersection between this two intervals are (-4, 3). answer: – 4 \leqslant x \leqslant 3.

Therefore, the solution from the two cases are $- 4 \leqslant 𝑥 \leqslant 3$ or in other words $(- 4, 3)$.

Example 2:- Solve $2x^2 + x – 3 < 0$. First factorize in the form of $m.n < 0$.

Solution: to factorize $2x^2 + x – 3 < 0$ find the two numbers p and q. where, $p + q = 1$, $p.q = 2(-3) = – 6$.

Thus, p and q can be 3 and – 2. and the middle term 1𝑥 is written as – 2x + 3x.

$2x^2 + x – 3 < 0$ will be $2x^2 – 2x + 3x – 3 < 0$, it can be grouped $(2x^2 – 2x) + (3x – 3) < 0$. take the common terms as $2x(x – 1) + 3 (x – 1) < 0$. also take the common terms as $(x – 1) (2x + 3) < 0$.

According to product property 2, to be

𝒎 ∙ 𝒏 < 𝟎 if and only if i. 𝒎 > 𝟎 and 𝒏 < 𝟎

case i. $x – 1 < 0$ and $2x + 3 > 0$

x < 1 and x > $\frac{-3}{2}$ . Thus, the intersection between this two intervals are ($\frac{-3}{2}$, 1). answer: $\frac{-3}{2} < x < 1$.

or case ii. $x – 1 > 0$ and $2x + 3 < 0$

$x > 1$ and $x < \frac{-3}{2}$. Thus, there are no any intersection between this two intervals. Answer: no solution.

Therefore, the solution from the two cases are $\frac{-3}{2} < x < 1$ or in other words $(\frac{-3}{2}, 1)$.

Example 3: Solve $(x + 4)(x – 3) \geqslant 0$?

Solution: According to product property 1, to be

m ∙ n > 𝟎 if and only if i. m > 𝟎 and n > 𝟎 or ii. m < 𝟎 and n < 𝟎

case i. $x + 4 \geqslant 0$ and $x – 3 \geqslant 0$

$x \geqslant – 4$ and $x \geqslant 3$ . Thus, the intersection of this two intervals are $x \geqslant 3$. Answer: $(3, \infty )$.

or case ii. $x + 4 \leqslant 0$ and $x – 3 \leqslant 0$

$x \leqslant – 4$ and $x \leqslant 3$. Thus, the intersection of this two intervals are $x \leqslant – 4$. answer: $(-\infty, -4 )$.

Therefore, the solution from the two cases are $x \leqslant – 4$ or $x \geqslant 3$ or in other words $(-\infty, -4 ) \cup (3, \infty )$.

Example 4:– Solve $2x^2 + x – 3 > 0$. First factorize in the form of $m.n < 0$.

Solution: to factorize $2x^2 + x – 3 > 0$ find the two numbers p and q. where, $p + q = 1$, $p.q = 2(-3) = – 6$.

Thus, p and q can be 3 and – 2. and the middle term 1𝑥 is written as – 2x + 3x.

$2x^2 + x – 3 > 0$ will be $2x^2 – 2x + 3x – 3 > 0$, it can be grouped $(2x^2 – 2x) + (3x – 3) > 0$. take the common terms as $2x(x – 1) + 3 (x – 1) > 0$. also take the common terms as $(x – 1) (2x + 3) > 0$.

According to product property 1, to be

m ∙ n > 𝟎 if and only if i. m > 𝟎 and n > 𝟎 or ii. m < 𝟎 and n < 𝟎

case i. $x – 1 > 0$ and $2x + 3 > 0$

x > 1 and x > $\frac{-3}{2}$ . Thus, the intersection of the two intervals are $x > 1$. answer: $(1, \infty )$.

or Case ii. $x – 1 < 0$ and $2x + 3 < 0$

$x < 1$ and $x < \frac{-3}{2}$. Thus, the intersection between this two intervals is $x < \frac{-3}{2}$. Answer: $(-\infty , \frac{-3}{2})$.

Therefore, the solution from the two cases are $x > 1$ or $x < \frac{-3}{2}$ in other words $(1, \infty )$ or $(-\infty , \frac{-3}{2})$.

4.2 Solving Quadratic Inequalities Using Sign Chart

Steps in solving quadratic inequalities by sign chart:

1. Re-write the inequality to get a 0 on the right-hand side.

2. Factor (if possible) the left-hand side.

3. Identify the roots or zeros of the factors obtained from step 2. These critical points divide the number line into intervals.

4. Choose a test point from each interval and evaluate the sign of each factor at that point.

5. Create a sign chart by listing the intervals and the signs of the factors in each interval.

6. Determine the sign of the product of the factors in each interval based on the signs obtained in step 5.

7. Identify the intervals where the product is positive or negative, which will give you the solution to the quadratic inequality.

8. Express the solution set in interval notation.

Example 5: Solve $(x + 4)(x – 3) \leqslant 0$

Solution: the first two steps are done.

step 3: Identify the roots or zeros of the factors x + 4 = 0, gives x = – 4. and x – 3 = 0 gives x = 3.

Prepare Sign chart table

	$x \le – 4$	$x = – 4$	$- 4 < x < 3$	$x = 3$	$x \geqslant 3$
x + 4	—	0	+++	+++	+++
x − 3	—	—	—	0	+++
$(x + 4)(x – 3)$	+++	0	—	0	+++

As the sign chart table shows.

The solution of $(x + 4)(x – 3) \leqslant 0$ is $- 4 < x < 3$ or (- 4, 3).

From the chart the solution of $(x + 4)(x – 3) \geqslant 0$ is $x \leqslant – 4$ or $x \geqslant 3$ or in other words $(-\infty, -4 ) \cup (3, \infty )$.

Example 6:- Solve $2x^2 + x – 3 \geqslant 0$. First factorize in the form of $m.n < 0$.

Solution: to factorize $2x^2 + x – 3 \geqslant 0$ find the two numbers p and q. where, $p + q = 1$, $p.q = 2(-3) = – 6$.

Thus, p and q can be 3 and – 2. and the middle term 1𝑥 is written as – 2𝑥 + 3𝑥.

$2x^2 + x – 3 \geqslant 0$ will be $2x^2 – 2x + 3x – 3 \geqslant 0$, it can be grouped $(2x^2 – 2x) + (3x – 3) \geqslant 0$. take the common terms as $2x(x – 1) + 3 (x – 1) \geqslant 0$. also take the common terms as $(𝑥 – 1) (2𝑥 + 3) \geqslant 0$.

the first two steps are done.

step 3: Identify the roots or zeros of the factors $𝑥 – 1 = 0$, gives $𝑥 = 1$. and $2𝑥 + 3 = 0$ gives $𝑥 = \frac{-3}{2}$.

Prepare Sign chart table

	$ x \leqslant \frac{- 3}{2}$	$ x = \frac{-3}{2} $	$ \frac{-3}{2} < x < 1 $	$ x = 1 $	$ x \geqslant 1 $
x – 1	—	—	—	0	+++
2x + 3	—	0	+++	+++	+++
$(x + 4)(x – 3)$	+++	0	—	0	+++

As the sign chart table shows.

The solution of $2x^2 + x – 3 \geqslant 0$ is $x \leqslant \frac{-3}{2}$ or $x \geqslant 1$ or in other words $(-\infty, \frac{-3}{2}) \cup (1, \infty)$.

From the chart the solution of $2x^2 + x – 3 \geqslant 0$ is $\frac{-3}{2} < x < 1$ or in other words $\left[ \frac{-3}{2}, 1 \right]$.

Applications

Example 7:

1.A recipe calls for at most 200 grams of sugar, but you have a mixture of 100 grams of sugar and another 150 grams of sugar substitute. How much of the sugar substitute can you use without exceeding the 200 grams limit?

2. You are trying to maintain your daily calorie intake between 1800 and 2200 calories. If your breakfast is 400 calories and lunch is 700 calories, how many calories can you have for dinner?

3. You want to study at least 15 hours a week but not more than 25 hours to balance other activities. How many hours should you study each day if you study 5 days a week?

3. How can quadratic inequalities be used to model habitat area constraints in ecological conservation efforts?

Solution:
In ecological conservation, preserving sufficient habitat area is crucial for the survival of species. If the habitat area ( A ) is modeled by a quadratic function due to varying environmental factors, conservationists need to ensure that this area remains within a certain range to support species effectively. For example, if the habitat area should not fall below a minimum threshold, the quadratic inequality $ax^2 + bx + c \geq M$ can be used.

Application: By solving the quadratic inequality $ax^2 + bx + c \geq M$, conservationists can identify the conditions under which the habitat area meets the minimum requirement. For instance, if the habitat area function is $-2x^2 + 10x + 20 \geq 15 $, solving this inequality helps determine the range of environmental variables that maintain the habitat area above the minimum threshold.

These questions illustrate how quadratic inequalities are used to model and solve real-life problems in various fields, from engineering to ecology, by applying mathematical concepts to practical constraints and optimization issues.

4. How can quadratic inequalities be applied to assess the safety of a vehicle’s braking distance under different conditions?

Solution
Vehicle safety often involves analyzing the braking distance, which can be modeled by a quadratic function of the vehicle’s speed. Suppose the braking distance ( D ) is given by $ D = ax^2 + bx + c $, where ( x ) is the initial speed of the vehicle. To ensure safety, the braking distance should not exceed a certain threshold ( T ). The quadratic inequality $ax^2 + bx + c \leq T$ represents this safety constraint.

Application: By solving the quadratic inequality $ax^2 + bx + c \leq T$, engineers can determine the maximum initial speed at which the vehicle can travel without exceeding the safe braking distance. For instance, if the braking distance function is $0.5x^2 + 2x + 10 \leq 100$, solving this helps find the maximum speed for safe braking.

Applications of Inequalities

Applications of Inequalities: Using inequalities to model and solve real-world problems involving limits, constraints, and ranges, such as budgeting, optimization, and determining feasible solutions.

Example 1. Budgeting and Finances

Scenario: Saving for a Phone

• Problem: You want to buy a phone that costs $600$ birr. You save $50$ birr per week. How many weeks do you need to save at least $600$ birr?
• Inequality: Let $w$ represent the number of weeks. $50w \geq 600$
• Solution: Divide both sides by 50 to find $w$. $w \geq 12$
• Interpretation: You need to save for at least 12 weeks to afford the phone.

Example 2. Cooking and Recipes

Scenario: Adjusting a Recipe

• Problem: A recipe calls for at most 200 grams of sugar, but you have a mixture of 100 grams of sugar and another 150 grams of sugar substitute. How much of the sugar substitute can you use without exceeding the 200 grams limit?
• Inequality: Let x represent the amount of sugar substitute used. $100 + x \leq 200$
• Solution: Subtract 100 from both sides to solve for x. $x \leq 100$
• Interpretation: You can use up to 100 grams of the sugar substitute.

Example 3. Sports and Fitness

Scenario: Running a Race

• Problem: You are training for a race and want to run more than 20 miles per week but no more than 35 miles to avoid injury. What is the range of miles you should run each week?
• Inequality: Let m represent the miles run per week. $20 < m \leq 35$
• Interpretation: You should run more than 20 miles but no more than 35 miles each week.

Example 4. Shopping and Discounts

Scenario: Buying Clothes

• Problem: You have $100$ birr to spend on clothes. You want to buy a pair of jeans that costs $40$ birr and spend the rest on t-shirts. Each t-shirt costs $15$ birr. How many t-shirts can you buy?
• Inequality: Let t represent the number of t-shirts. $40 + 15t \leq 100$
• Solution: Subtract 40 from both sides and then divide by 15. $15t \leq 60 \implies t \leq 4$
• Interpretation: You can buy up to 4 t-shirts.

Example 5. Travel and Transportation

Scenario: Car Rental

• Problem: A car rental company charges $50$ birr per day plus a one-time fee of $20$ birr. You have a budget of $200$ birr. How many days can you rent the car without exceeding your budget?
• Inequality: Let d represent the number of days. $50d + 20 \leq 200$
• Solution: Subtract 20 from both sides and then divide by 50. $50d \leq 180 \implies d \leq 3.6$
• Interpretation: You can rent the car for up to 3 days.

Example 6. Nutrition and Diet

Scenario: Calorie Intake.

• Problem: You are trying to maintain your daily calorie intake between 1800 and 2200 calories. If your breakfast is 400 calories and lunch is 700 calories, how many calories can you have for dinner?
• Inequality: Let c represent the calories for dinner. 1$1800 \leq 400 + 700 + c \leq 2200$
• Solution: Subtract 1100 from all parts of the inequality. $700 \leq c \leq 1100$
• Interpretation: You can have between 700 and 1100 calories for dinner.

Example 7. Academic Goals

Scenario: Studying for Exams

• Problem: You want to study at least 15 hours a week but not more than 25 hours to balance other activities. How many hours should you study each day if you study 5 days a week?
• Inequality: Let h represent the hours per day. $15 \leq 5h \leq 25$
• Solution: Divide all parts by 5. $3 \leq h \leq 5$
• Interpretation: You should study between 3 and 5 hours each day.