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Comprehensive Guide to Solving Quadratic Inequalities in One Variable

Learn to solve quadratic inequalities using factoring and the quadratic formula, interpret solutions on number lines, and express answers in inequality, interval, and set notations.

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QUADRATIC INEQUALITIES IN ONE VARIABLE Math 10, Term 1, Week 5
LEARNING COMPETENCIES 1. illustrate quadratic inequalities in one variable in the number line. 2. solve quadratic inequalities in one variable and express solutions in various notations.
DAY
Short review
WRITE Q IF THE EQUATION IS QUADRATIC AND NQ IF IT IS NOT QUADRATIC. NQ Q NQ Q Q 9. 10. NQ Q NQ Q Q ANSWER KEY
ESSENTIAL QUESTIONS 1.How many correct answers did you get? 2.What did you find easy to do in the activity? 3.What did you find difficult? 4.What techniques made the activity easy? 5.How do you feel about the result?
LESSON PURPOSE
It is true that we, humans, are created equal. However, in real life, we are dealing with unfair situations like one family is better than others, one school is better than our school, or we are less privileged than others, etc. This means that INEQUALITY is real
From solving quadratic equations, are you now ready to solve quadratic INEQUALITIES?
Did you know? Quadratic inequalities can be used to model situations like … firing a cannon
Did you know? Quadratic inequalities can be used to model situations like … Hitting a baseball/Golf Ball
UNLOCKING CONTENT VOCABULAR
Quadratic Inequalities are similar to quadratic equations only that they use inequality symbols (<, >, , ≤ x2 – 3x + 5 = 0 quadratIc InequalIty < > ≤ ≥
When plotted as graphs, they display a parabola. quadratIc InequalIty
The standard forms of a quadratic inequality are: where a, b, and c are real numbers and a ≠ 0. QUADRATIC INEQUALITY 1) ax2 + bx + c < 0 2) ax2 + bx + c > 0 3) ax2 + bx + c 0 ≥ 4) ax2 + bx + c 0 ≤
DAY
Short review
Solve each quadratic equation by factoring 1) x2 – 2x = 0 2) x2 + 2x – 3 = 0 3) x2 – x – 6 = 0 4) x2 – 5x + 6 = 0 5) 2x2 + 5x – 3 = 0 x = 0, 2 x = 1, –3 x = 3, –2 x = 3, 2 x = 1, –6 ANSWER KEY 6) x2 = 4 7) x2 – 2x = 15 8) x2 = 3x – 2 9) 3x2 – 12 = 5x 10) 3x2 = x + 24 x = 2, –2 x = 5, –3 x = 1, 2 x = 3, –4 /3 x = 3, –8 /3
SOLVING QUADRATIC INEQUALITIES IN ONE VARIABLE BY FACTORING SUB-TOPIC 1
EXPLICITAT ION
QUADRATIC INEQUALITY IN ONE VARIABLE Quadratic Inequalities are similar to quadratic equations only that they use inequality symbols (<, >, , ) ≤ ≥ instead of an equal sign (=).
QUADRATIC INEQUALITY IN ONE VARIABLE To solve quadratic inequalities, like quadratic equations, one needs to be familiar with factoring techniques but instead of two solutions, there are a range of solutions (intervals). x2 – 3x – 10 > 0 x < -2 or x > 5 (−∞,−2) (5,∞) ∪ ​ x2 – 3x – 10 = 0 x = 5, x = -2 {5, -2} QE QI
Quadratic Inequality in One Variable Another important similarity of quadratic equations and inequalities is the use of the Zero-Factor property (ab = 0 if a = 0 or b = 0 or both.). The product of any number and zero is always zero
he steps on how to solve quadratic inequalities by factoring are: 1) Set one side of the inequality to zero (standard form). 2) Factor the quadratic expression. 3) Equate each (linear) factor to zero to find the critical points. 4) Create intervals from the critical points. 5) Use Sign Test for each interval on the factored inequality using a test point. 6) Write the solution that makes the inequality true using the inequality notation, interval notation, or set notation.
Recall: 1) If a number x > 0, this means that the number x is POSITIVE. x > 0 {1, 2, 3, 4, 5, 6, …} 2) If a number x < 0, this means the number x is NEGATIVE. x > 0 {-1, -2, -3, -4, -5, -6, …}
WORKED EXAMPLE
Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Solution: Step 1: Observe that one side of the inequality is already zero. Step 2: Factor: x2 + 4x – 5 = (x + 5) (x – 1). Step 3: Equate: Critical points: Step 4: Create Intervals: x + 5 = 0 x - 1 = 0 x = -5 x = 1 x < -5, x > 1 –5 < x < 1, x < -5 –5 < x < 1 x > 1
Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Step 5: Use Sign Test: interval/s that make (x + 5)(x – 1) < 0 TRUE. For x < –5, test x = –6 (x + 5)(x – 1)< 0 (-6+5)(-6–1) < 0 (–1)(–7) < 0 +7 < 0 FALSE For –5 < x < 1, test x = 0 (x+5)(x–1) < 0 (0+5)(0–1) < 0 (5)(–1) < 0 -5 < 0 TRUE For x > 1, test x = 3 (x+5)(x–1)< 0 (3+5)(3–1) < 0 (8)(2) < 0 16 < 0 FALSE
Factors (–∞,–5) Test: x = –6 (–5, 1) Test: x = 0 (1, +∞) Test: x = 3 (x + 5) - + + (x – 1) - - + (x + 5)(x – 1) + - + True/False False True False Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). or
Step 6: The solution using inequality notation is –5 < x < 1 or (–5, 1) if in interval notation. Figure 2. Graph of y = x2 + 4x – 5 and its solution (–5, 1) Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Note: (–5, 1) • means all numbers between -5 and 1, but not including -5 and 1 • means {-4, -3.5, -3, -2, -1.25 -1, 0.5. 0}
Example 2. Solve x2 + 6x > –8. Solution: Step 1: Set one side of the inequality to zero: x2 + 6x + 8 > 0. (POSITIVE Step 2: Factor: x2 + 6x + 8 = (x + 4)(x + 2). Step 3: Equate: Critical points: Step 4: Create Intervals: x+4=0 x+2=0 x=-4 x=-2 x < -4 x > -2 –4< x <-2, x < -4 –4<x<-2 x > -2
Step 5: Use Sign Test: interval/s that make (x + 4)(x + 2) > 0 TRUE. For x < –4, test x = –5 (x+4)(x 2) > 0 (-5+4)(-5+2) > 0 (-1)(–3) > 0 3 > 0 TRUE For –4 < x < -2, test x = -3 (x+4)(x+2) > 0 (-3+5)(-3–1) > 0 (2)(–4) > 0 -8 > 0 FALSE For x > -2, test x = 0 (x+4)(x+2) > 0 (0+4)(0+2) > 0 (4)(2) > 0 8 > 0 TRUE Example 2. Solve x2 + 6x + 8 > 0
Factors (–∞,–4) Test: x = –6 (–4, -2) Test: x = 0 (-2, +∞) Test: x = 3 (x + 4) - + + (x + 2) - - + (x + 4)(x + 2) + - + True/False True False True Example 2. Solve x2 + 6x + 8 > 0 or
Step 6: The solution of the inequality is x < –4 or x > –2 in inequality notation, (– , –4) U (–2, + ) in interval notation, ∞ ∞ or {x|x < –4 or x > –2} in set notation Note: (– , –4) U (–2, + ) ∞ ∞ • means -4, -3, -2 are not included in the solution • means -5, -7.5, -20, -1, 5.25 -1, 0.5. 12.65, … Example 2. Solve x2 + 6x + 8 > 0 x < -4 –4<x<-2 x > -2
Example 3. Solve 2x2 11x – 12 ≥ Solution: Step 1: Set one side of the inequality to zero: 2x2 – 11x + 12 0 (+) ≥ Step 2: Factor: Factor: 2x2 – 11x + 12 = (x – 4)(2x – 3). Step 3: Equate: Critical points: Step 4: Create Intervals: x - 4 = 0 2x - 3 = 0 x = 4 x = 3 /2 or 1.5 x ≤ 1.5, x 4 ≥ 1.5 x 4, ≤ ≤ Step 6: The solution is x 1.5 or x 4 ≤ ≥ OR (– , 1.5] U [4, + ). ∞ ∞ That means 2, 2.5, 3 are not included in the solution x 1.5 ≤ 1.5≤x≤ 4 x 4 ≥ Step 5. Use Sign Test: interval/s that make (x – 4)(2x – 3). 0 TRUE ≥ For 1.5 x 4, ≤ ≤ test x = 3 (x - 4)(2x - 3) 0 ≥ (3 - 4)(2(3) - 3) 0 ≥ (-1)(3) 0 ≥ -2 0 ≥ FALSE
LESSON ACTIVITY
EXERCISE 1 SOLVE THE FOLLOWING QUADRATIC INEQUALITIES BY FACTORING USING THE 6 STEPS. 1) x2 + 5x + 6 > 0 2) x2 – x < 6 3) 3x2 > 2x + 5 4) 2x2 + 5x ≤ 3 5) 5x2 > 8x – 3 x < –3 or x > –2 OR (–∞,–3) U (–2, +∞) –2 < x < 3 OR (–2, 3) x < –1 or x > 5 /3 OR (–∞,–1) U (5 /3, +∞) –3 ≤ x ≤ ½ OR [-3, ½] x < 3 /5 or x > 1 OR (–∞, 3 /5) U (1, +∞) ANSWER KEY
DAY
Short review
Complete the steps in solving 𝑥2 + 5 + 6 = 0 using the 𝑥 QF 1. Identify , , and . 2. Write the quadratic formula. 3. Substitute the values. 5. Find the square root. 6.Solve for the two values of . Final Answer: 1 5 6 5 5 1 6 1 25 1 = 1 -2 -3 -2 -3
SOLVING QUADRATIC INEQUALITIES IN ONE VARIABLE BY THE QUADRATIC FORMULA SUB-TOPIC 2
EXPLICITAT ION
QUADRATIC INEQUALITY IN ONE VARIABLE When solving the quadratic inequality by factoring does not work, like quadratic equations, the next best thing to do is use the quadratic formula. We still use the same steps, but instead of factoring, we find the critical points using the quadratic formula (Steps 2 & 3).
QUADRATIC INEQUALITY IN ONE VARIABLE Recall: For the quadratic equation in standard form ax2 + bx + c = 0,
WORKED EXAMPLE
Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Solution: Step 1: Observe that one side of the inequality is already zero. Step 2: Standard form: x2 + 4x – 5 = 0 where a=1, b=4, c=-5 Step 3: Plug in the values in the formula: = = = = = = 1 = = -5 Step 4: Create Intervals: x < -5, x > 1 –5 < x < 1, x < -5 –5 < x < 1 x > 1
Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Step 5: Use Sign Test: interval/s that make (x + 5)(x – 1) < 0 TRUE. For x < –5, test x = –6 (x + 5)(x – 1)< 0 (-6+5)(-6–1) < 0 (–1)(–7) < 0 +7 < 0 FALSE For –5 < x < 1, test x = 0 (x+5)(x–1) < 0 (0+5)(0–1) < 0 (5)(–1) < 0 -5 < 0 TRUE For x > 1, test x = 3 (x+5)(x–1)< 0 (3+5)(3–1) < 0 (8)(2) < 0 16 < 0 FALSE
Factors (–∞,–5) Test: x = –6 (–5, 1) Test: x = 0 (1, +∞) Test: x = 3 (x + 5) - + + (x – 1) - - + (x + 5)(x – 1) + - + True/False False True False Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). x < -5 –5 < x < 1 x > 1 or
Step 6: The solution using inequality notation is –5 < x < 1 or (–5, 1) if in interval notation. Figure 2. Graph of y = x2 + 4x – 5 and its solution (–5, 1) Example 1. Solve x2 + 4x – 5 < 0 (NEGATIVE). Note: (–5, 1) • means all numbers between -5 and 1, but not including -5 and 1 • means {-4, -3.5, -3, -2, -1.25 -1, 0.5. 0}
LESSON ACTIVITY
EXERCISE 1 SOLVE THE FOLLOWING QUADRATIC INEQUALITIES BY FACTORING USING THE 6 STEPS. 1) x2 + 5x + 6 > 0 2) –x2 – 6x > 5 3) –x2 + 10x ≤ 16 4) 2x2 + 5x ≤ 3 5) x2 ≤ 2x + 1 x < –3 or x > –2 OR (–∞,–3) U(–2, +∞) –5 < x < –1 OR (–5, –1) x ≤ 2 or x ≥ 8 OR (–∞, 2] U [8, +∞) –3 ≤ x ≤ ½ OR [-3, ½] 1−√2≤ ≤1+√2 ] OR [1−√2, 1+√2] 𝑥 ANSWER KEY
DAY
Short review
x2 − 5x + 6 > 0 1. Which of the following is the correct factor of x2 − 5x + 6? A. (x+2)(x+3) C. (x+1)(x−6) B. (x−2)(x−3) D. (x−1)(x−6)
x2 − 5x + 6 > 0 2. Based on your answer in Question 1, what are the values of x? A. x = −2, −3 B. x = 2, 3 C. x = −1, 6 D. x = 1, 6
x2 − 5x + 6 > 0 3. Which inequalities show the intervals that should be tested? A. x<2, 2<x<3, x>3 B. x<−2, −2<x<−3, x>−3 C. x<3, x>3 D. x<2, x>2
x2 − 5x + 6 > 0 4. If x = 2.5 is substituted in the inequality (x−2)(x−3) > 0, is the statement true or false? A. True B. False (2.5−2)(2.5−3) > 0 (0.5)(−0.5) > 0 −0.25 > 0
x2 − 5x + 6 > 0 5. What is the solution set of x2 −5x+6 > 0? A. (2,3) B. [2,3] C. (−∞,2) (3,∞) ∪ D. (−∞,2] [3,∞) ∪
PROBLEMS INVOLVING QUADRATIC INEQUALITIES IN ONE SUB-TOPIC 3
EXPLICITAT ION
QUADRATIC INEQUALITY IN ONE VARIABLE Some problems in the sciences and economics involve quadratic inequalities. Quadratic inequalities can be used to model situations like firing and shooting a cannon or hitting a baseball or golf ball among other applications.
WORKED EXAMPLE
Example 1 The length of a rectangular field is 10 meters more than twice its width. Find all possible measure of the width that will result in the area of the rectangular field not exceeding 100 square meters. width length A ≤ 100m2
Example 1 The length of a rectangular field is 10 meters more than twice its width. Find all possible measure of the width that will result in the area of the rectangular field not exceeding 100 square meters. Solution: Let w represent the width and l represents the length of the field. Given the information in the first sentence of the problem, we have l = 2w + 10 Then from the formula of the area of a rectangle; Area = length x width A = l • w Substituting 2w + 10 for l in the formula of the area, we have A = l • w = (2w + 10) • w = 2w2 + 10w Therefore: A = 2w2 + 10w
Example 1 The length of a rectangular field is 10 meters more than twice its width. Find all possible measure of the width that will result in the area of the rectangular field not exceeding 100 square meters. Solution: From the problem, it says the area cannot exceed 100 m2 , so 2w2 + 10w ≤ 100, a quadratic inequality. From here on, we can follow the steps described earlier. Step 1: One side of the inequality is zero: 2w2 + 10w – 100 ≤ 0. Step 2: Factor 2w2 + 10w – 100 = 2(w2 + 5w – 50) = 2(w + 10) (w – 5) Step 3: Equate w + 10 = 0 w – 5 = 0 Critical points: w = -10 w = 5
Example 1 Solution: Step 4: Intervals: Since width cannot be negative, we can safely ignore -10. So the intervals to consider are w ≤ 5 and w ≥ 5. Step 5: Sign Test: interval/s that make 2(w + 10) (w – 5) ≤ 0 TRUE. For w ≤ 5, test w = 3: 2(w + 10)(w – 5) ≤ 0 2(3 + 10)(3 – 5) ≤ 0 2(13)(-2) ≤ 0 -52 ≤ 0 TRUE For w ≥ 5, test w = 6: 2(w + 10)(w – 5) ≤ 0 2(6 + 10)(6 – 5) ≤ 0 2(16)(1) ≤ 0 32 ≤ 0 FALSE
Example 1 Solution: Step 6: The solution is w ≤ 5 or the width of the rectangular field must be less than or equal to 5 meters. The possible width could be any number less than or equal to 5 meters but cannot be a negative number nor 0. w = {1, 1.25, 1.5, 2, 3, 4, 4.75, 5}
LESSON ACTIVITY
EXERCISE 1 Solve the following problems involving quadratic inequality. 1) The length of a rectangular court is 9 feet more than twice its width. Find all possible measure of the width that will result in the area of the rectangular court not exceeding 200 square feet. 2) An actor of a movie will jump off from the top of a building 20 meters high. A high-speed camera is being prepared to capture the actor between 15meters and 10 meters above the ground. How long after the jump does the cameraman film him? (Hint: use this formula simplified from physics d = 20 – 5t2 where d = distance above ground (m) and t = time, in seconds, from jump off. Also note that 10 < d < 15)
ANSWER KEY 1) The width must be greater than 0 feet but not more than 8 feet. 2) The cameraman films the actor from after 1 second until before about 1.41 seconds after the jump
Comprehensive Guide to Solving Quadratic Inequalities in One Variable