In this lesson, we will focus on quadratic equations and inequalities and algebraic methods to solve them.

Lesson 7 Quadratic Equations, Inequalities, and Factoring Lesson 7 Quadratic Equations, Inequalities and Factoring In this lesson, we will focus on quadratic equations and inequalities and algebraic methods to solve them. We will begin by learning what it means to factor and then we will apply specific types of factoring to break apart quadratic expressions. We will end this lesson by learning how to use factoring to solve quadratic equations and then we will apply factoring methods to help us solve quadratic inequalities. Lesson Topics Section 7.1: Factoring Quadratic Expressions What Does it Mean to Factor? Factoring Using the Method of Greatest Common Factor (GCF) Factoring by Trial and Error Special Forms Prime Quadratic Expressions Section 7.2: Solving Quadratic Equations by Factoring Section 7.3: Solving Quadratic Inequalities by Factoring Section 7.4: Applications of Quadratic Inequalities 277

Lesson 7 Quadratic Equations, Inequalities, and Factoring 278

Lesson 7 - MiniLesson Section 7.1 Factoring Quadratic Expressions So far, we have only used our graphing calculators to solve quadratic equations utilizing the Graphing/Intersection Method. There are other methods to solve quadratic equations. The first method we will discuss is the method of factoring. Before we jump into this process, you need to have some concept of what it means to factor using just numbers. Factoring Whole Numbers To factor 60, there are many responses some of which are below: 60 = 1 60 (not!very!interesting!but!true) 60 = 2 30 60 = 3 20! 60 = 4 3 5 All of these factorizations of 60 rewrite 60 as products of some of the numbers that divide it evenly. The most basic factorization of 60 is as a product of its prime factors. Remember that prime numbers are only divisible by themselves and 1. The prime factorization of 60 is:!60 = 2 2 3 5 There is only one prime factorization of 60 so we can now say that 60 is completely factored when we write it as!60 = 2 2 3 5. When we factor quadratic expressions, we use a similar process. This process involves factoring expressions such as!24x 2. To factor!24x 2 completely, we would first find the prime factorization of 24 and then factor!x 2.!24 = 2 2 2 3 and!x 2 = x x Putting these factorizations together, we obtain the following:!24x 2 = 2 2 2 3 x x Let us see how the information above helps us to factor more complicated quadratic expressions and ultimately leads us to a second solution method for quadratic equations. 279

Problem 1 WORKED EXAMPLE Factoring Quadratic Expressions Using GCF Method Completely factor the quadratic expression! 3x 2 + 6x. Check your final result. Completely factor the terms!3x 2 and!6x individually to obtain what you see below.! 3x 2 =!3 x x!!!!!!and!!!!6x =!3 2 x Let s rearrange these factorizations just slightly as follows: We can see that! 3 x factor (GCF) of both terms.! 3x 2 = ( 3 x) x!!!!!!and!!!!6x = ( 3 x) 2 ( ) = 3x is a common factor to both terms. In fact, 3x is the greatest common Let s rewrite the full expression with the terms in factored form and see how that helps us factor the expression: 3x 2 + 6x = (3 x) x + (3 x) 2 = (3x) x + (3x) 2 = (3x)(x + 2) = 3x(x + 2) Result:! 3x(x + 2) is completely factored form for! 3x 2 + 6x Always check your factorization by multiplying the final result. ( ) = 3x( x) + 3x( 2) =!3x 2 +!6x 3x x +!2! CHECKS Problem 2 MEDIA EXAMPLE Factoring Quadratic Expressions Using GCF Method Completely factor the quadratic expressions below. Check your final results. a)! 11a2 4a b)!55w 2 +!5w 280

Problem 3 YOU TRY Factoring Quadratic Expressions Using GCF Method Completely factor the quadratic expressions below. Check your final results. a) 64b 2 16b b) 11c 2 + 7c Quadratic expressions with 3 terms are called trinomials. The Trial and Error Method to factor trinomials is shown below. Factoring Quadratic Expressions of the form! x 2 + bx + c by Trial and Error! x 2 + bx + c = ( x! +!p)( x! +!q), where b! =!p! +!q and c! =!p q Problem 4 WORKED EXAMPLE Factoring Quadratic Expressions Using Trial and Error Completely factor the quadratic expression! x 2 +!5x!6. Check your final result. Step 1: Look to see if there is a common factor in this expression. If there is, then use the GCF method to factor out the common factor. The expression! x 2 +!5x!6 has no common factors. Step 2: For this problem, b = 5 and c = 6. We need to identify p and q. These are two numbers whose product is 6 and sum is 5. A helpful method to identify p and q is to list different numbers whose product is 6 and then see (i.e. trial and error) which pair adds to 5. Product = 6 Sum = 5 3 2 No 3 2 No 1 6 YES 1 6 No Step 3: Write in factored form x 2 + 5x 6 = (x + ( 1))(x + 6) x 2 + 5x 6 = (x 1)(x + 6) Step 4: Check by foiling. (x 1)(x + 6) = x 2 + 6x x 6 = x 2 + 5x 6 CHECKS! 281

Problem 5 MEDIA EXAMPLE Factoring Quadratic Expressions Using Trial and Error Completely factor the following quadratic expressions. Check your final results. a) a 2 + 7a + 12 b) w 2 + w 20 c) x 2 36 Problem 6 YOU TRY Factoring Quadratic Expressions Using Trial and Error Completely factor the following quadratic expressions. Check your final results. a)! n2 + 8n + 7 b)!r 2 +3r 70 c)! m2 m 30 d)! x 2 5x 24 e)! r 2 + 2r 35 f)! x 2 + x 6 282

In previous examples, the coefficient of the squared term, a, was 1. When! a 1, the same tools we have learned up to this point will apply. Problem 7 WORKED EXAMPLE Factoring Quadratic Expressions using Trial and Error! a 1 Completely factor the quadratic expression,! 2x 2 7x 4. Check your result. Step 1: Look to see if there is a common factor in this expression. If there is, then use the GCF method to factor out the common factor. The expression! 2x 2 7x 4 has no common factors. Step 2: The only way we can get!2x 2 as the first term is to multiply!2x times!x. Also, because the constant term is negative (! 4 ) our factoring signs must alternate. We just don t know which way. This means our factorization would look like one of these:! (2x + )(x ) or! (2x )(x + ) Step 3: We are looking for factors of -4 to go into the last spot for each set of ( ). Our choices are combinations of 4 and 1, or, 2 and 2. Let s try each combination to see what works. This is the trial and error part. (2x + 4)(x 1) = 2x 2 2x + 4x 4 = 2x! 2 2x 4 Does not give us! 2x 2 7x!4 (2x 1)(x + 4) = 2x 2 + 8x x 4 =! 2x 2 + 7x 4 Does not give us! 2x 2 7x!4 (2x + 2)(x 2) = 2x 2 4x + 2x 4 =! 2x 2 2x 4 Does not give us! 2x 2 7x!4 (2x + 1)(x 4) = 2x 2 8x + x 4 =! 2x 2 7x 4 VOILA!!! This combination works! (2x 4)(x + 1) = 2x 2 + 2x 4x 4 =! 2x 2 2x 4 Does not give us! 2x 2 7x!4 Step 4: Write the final result in factored form:! 2x 2 7x 4 = (2x + 1)(x 4). We checked already using Trial and Error above so we do not need to recheck. Once we find one solution, we can stop as there will be only one way to completely factor. 283

Problem 8 MEDIA EXAMPLE Factoring Quadratic Expressions using Trial and Error! a 1 Completely factor the following quadratic expressions. Check your final results. a)! 3x 2 + x 14 b)! 2x 2 + 11x 5 Problem 9 YOU TRY Factoring Quadratic Expressions using Trial and Error! a 1 Completely factor the following quadratic expressions. Check your final results. a)! 3x 2 13x + 4 b)! 5x 2 7x 6 284

The problems that we work get more and more complicated depending on the coefficients involved. If there are more possibilities for factors of the a coefficient, then we have to work a little harder to factor. Problem 10 MEDIA EXAMPLE Factoring Quadratic Expressions,! a 1, more complicated Completely factor the following quadratic expressions. Check your final results. a)! 6x 2 5x 4 b)! 15x 2 + 8x 12 Problem 11 YOU TRY Factoring Quadratic Expressions,! a 1, more complicated Completely factor the following quadratic expressions. Check your final results. a)! 8x 2 + 14x 15 b)! 20x 2 44x 15 285

Going back to the method of greatest common factoring (GCF), there will be times when you first need to remove a greatest common factor prior to using the Trial and Error Method. See the following examples. Problem 12 MEDIA EXAMPLE Using multiple factoring methods: GCF and Trial/Error Completely factor the following quadratic expressions. Check your final results. a)! 2x 2 + 10x 12 b)! 6x 2 21x 12 Problem 13 YOU TRY Using multiple factoring methods: GCF and Trial/Error Completely factor the following quadratic expressions. Check your final results. a)! 3x 2 +!24x +!21 b)! 5t 2 5t 150 286

Special Quadratic Form Difference of Squares! a2 b 2 = (a b)(a+ b) Problem 14 WORKED EXAMPLE Special Quadratic Form Difference of Squares Completely factor the following quadratic expressions. Check your final results. a) x 2 9 = x 2 3 2 = (x 3)(x + 3) b) 4x 2 25 = (2x) 2 5 2 = (2x 5)(2x + 5) Check: Check: (x 3)(x + 3)= x 2 3x + 3x 9 (2x.5)(2x+5)=4x 2 +10x 10x 25 = x 2 9!!!!!!!!!!!!!!!!!!!!!!!!! = 4x 2 25 Final!Answer:! Final!Answer:! x 2 9 = (x 3)(x + 3)! 4x 2 25 = (2x 5)(2x +5) Problem 15 MEDIA EXAMPLE Special Quadratic Form Difference of Squares Completely factor the following quadratic expressions. Check your final results. a)! a2 16 b)! 2x 2 8 Check: Check: Problem 16 YOU TRY Special Quadratic Form Difference of Squares Completely factor the following quadratic expressions. Check your final results. a)! x 2 81 b)! 5x 2 5 Check: Check: 287

Problem 17 YOU TRY Factoring Quadratic Expressions Use the methods discussed in this section to completely factor the following quadratic expressions. Check your final results. a)! x 2 + 9x + 8 Check: b)! 2x 2 + 7x 4 Check: c)! 4a2 25 Check: d)! 6x 2 9x 15 Check: e)! 4x 2 13x 12 Check: 288

Prime Quadratic Expressions If the quadratic expression!ax 2 + bx + c cannot be factored using integer values, then we say that the quadratic expression is prime. Integers! = {... 3, 2, 1,0,1,2,3...} If you have tried all the factoring methods in this lesson and cannot find integer values that work, then you can say that the given quadratic expression is prime. Note that a prime quadratic is still factorable but the numbers will not be nice numbers like integers. The factors will be rational numbers or irrational numbers. Problem 18 WORKED EXAMPLE Prime Quadratic Expressions Verify that both of the following quadratic expressions are prime. To do that, we will use the Trial and Error Method and show that none of the combinations will work. a) a 2 + 9 =!Prime Option!1!Fails:! (a 3)(a 3)= a 2 6a + 9 Option!2!Fails:! (a + 3)(a + 3)= a 2 + 6a+ 9 b) 2x 2 2x 1 = Prime Option!1:!Fails ( 2x 1) x +1 2x +1! ( ) = 2x 2 + x 1 Option!2:!Fails ( )( x 1) = 2x 2 x 1 Problem 19 MEDIA EXAMPLE Prime Quadratic Expressions Verify that both of the following quadratic expressions are prime. a)! x 2 2x 1 b)! 4x 2 + 2x 5 Problem 20 YOU TRY Prime Quadratic Expressions Verify that! 4a2 + 25 is a prime quadratic expression. 289

Section 7.2 Solving Quadratic Equations by Factoring In this section, we will see how a quadratic equation written in standard form can be solved algebraically using factoring methods. The Zero Product Principle If a b = 0, then a = 0 or b = 0 To solve a quadratic equation by factoring: Step 1: Make sure the quadratic equation is in standard form: ax 2 + bx + c = 0. Step 2: Write the left side in completely factored form. Step 3: Apply the Zero Product Principal to set each linear factor = 0 and solve for x Step 4: Verify result by graphing and finding the intersection point(s). Problem 21 WORKED EXAMPLE Solve Quadratic Equations By Factoring a) Solve by factoring: 5x 2 10x = 0 Step 1: This quadratic equation is already in standard form. Step 2: Check if there is a common factor, other than 1, for each term Yes, 5x is common to both terms Step 3: The left side can be written in completely factored form using GCF method: 5x 2 10x = 0 5x(x 2) = 0 Step 4: Apply the Zero Product Principle to set each linear factor = 0 and solve for x: 5x = 0 or x 2 = 0 x = 0 or x = 2 Step 5: Verify result by graphing. Note that the solutions to 5x 2 10x = 0 are the x values of the horizontal intercepts seen in the graphs below. 290

b) Solve by factoring: x 2 7x + 12 = 2 Step 1: Make sure the quadratic is in standard form. Subtract 2 from both sides to get: x 2 7x + 10 = 0 Step 2: There is no common factor, other than 1, for each term. Step 3: Use trial and error to write the left side in completely factored form. x 2 7x + 10 = 0 (x 5)(x 2) = 0 Step 4: Set each linear factor to 0 and solve for x: (x 5) = 0 or (x 2)= 0 x = 5 or x = 2 Step 5: Verify result by graphing. Note that the solutions to x 2 7x + 12 = 2 are the x-values of the intersections seen in the graphs below. 291

Problem 22 MEDIA EXAMPLE Solve Quadratic Equations By Factoring Use an appropriate factoring method to solve each of the quadratic equations below. Show all of your work. Be sure to write your final solutions using proper notation. Verify your answer by graphing. Sketch the graph on a good viewing window (the vertex, vertical intercept, and any horizontal intercepts should appear on the screen). Mark and label the solutions on your graph. a) Solve by factoring:! 2x 2 = 8x b) Solve by factoring:! x 2 = 3x + 28 c) Solve by factoring:! 2x 2 5x + 4 = 2x + 9 292

Problem 23 YOU TRY Solving Quadratic Equations by Factoring Use an appropriate factoring method to solve each of the quadratic equations below. Show all of your work. Be sure to write your final solutions using proper notation. Verify your answer by graphing. Sketch the graph on a good viewing window (the vertex, vertical intercept, and any horizontal intercepts should appear on the screen). Mark and label the solutions on your graph. a) Solve! x 2 + 3x = 10 b) Solve!3x 2 = 15x c) Solve by factoring:! 2x 2 + 6x 1 = x + 2 293

Section 7.3 Solving Quadratic Inequalities by Factoring In this section, we will see how a quadratic inequality written in standard form can be solved algebraically using factoring methods. You should have studied linear inequalities previously but reviewing the basics is always a good idea. Problem 24 MEDIA EXAMPLE Review of Linear Inequalities and Notation Solve the following inequalities. Provide your final answers on a graph and also using inequality and interval notation. a)!x +5 0 b)!2x +12< 0 c)! 3x +6 0 When Does the Inequality Symbol Change Directions? Remember that when you multiply or divide by a negative number as part of your solution process, you must change the direction of the inequality symbol. When the Direction Changes In the example below, the inequality symbol changes direction when dividing both sides by! 3! 3x 12 3x 3 12 3 x 4 When the Direction Does Not Change In the example below, the inequality symbol does not change directions when subtracting 3 from both sides.! x + 3 12 x + 3 3 12 3 x 9 294

Quadratic Inequalities Quadratic inequalities are in one of the following standard forms:! ax 2 + bx + c > 0! ax 2 + bx + c < 0! ax 2 + bx + c 0! ax 2 + bx + c 0 Problem 25 WORKED EXAMPLE Solve Quadratic Inequalities By Factoring Solve the quadratic inequality by factoring:! 5x 2 10x > 0. Check by graphing. Step 1: This quadratic inequality is already in one of the standard forms. Step 2: Write the left side in completely factored form: 5x 2 10x > 0 5x(x 2) > 0 Step 3: Set each linear factor equal to 0 and solve for x: 5x = 0 OR x 2 = 0 x = 0 OR x = 2 Step 4: Place the values of x on a real number line and make a sign chart to determine the sign of 5x(x-2) in each of the corresponding segments: Step 5: The solution to this inequality lies in the segments above that are positive. Here is the solution in inequality and interval notations. Note that the symbol below means union and is the preferred notation to join together two intervals. Inequality Notation!x < 0!!or!!x > 2 Interval Notation! (,0) (2, ) 295

Step 6: Check the result by graphing. Graphing images and explanation are below. The graphing equations and window are listed, as is the resulting graph. The equations intersect at x = 0 and x = 2 which verifies our results from step 3. The inequality we began with was:! 5x 2 10x > 0 which asks us to look at outputs for the inequality that are strictly larger than 0. The x-axis denotes the output values that equal 0. Therefore we are interested in finding the x- values that give us y-values above the x-axis basically the arms of the parabola that are above the x-axis. When x < 0, the outputs are positive and when x > 2, the outputs are positive. This confirms our solutions from the algebraic approach. Inequality Notation!x < 0!!or!!x > 2 Interval Notation! (,0) (2, ) 296

Problem 26 MEDIA EXAMPLE Solve Quadratic Inequalities By Factoring Solve the quadratic inequality by factoring:!x 2!7x +!12!!2. Check by graphing. Leave your answer in interval and inequality notation. 297

Problem 27 YOU TRY Solve Quadratic Inequalities By Factoring Solve the quadratic inequality by factoring:!x 2 +!3x!10!!0. Check by graphing. Leave your answer in interval and inequality notation. 298

Section 7.4 Applications of Quadratic Inequalities Problem 28 MEDIA EXAMPLE Applications of Quadratic Inequalities The function! h(t)= 16t 2 +100t +3, where! h(t) is the height in feet after t seconds, models the path of a baseball hit from home plate into the playing field. a) Begin by drawing an accurate representation of the graph of! h(t) including: starting height, time to maximum height and maximum height and time to hit the ground. b) One day during a game, player Scott hit a ball whose path followed the model above. The stadium in which the game was played had a design flaw that prevented the fans in the middle sections from being able to view the ball during the times when its height was more than 100 feet. For how many seconds could the fans in this game not see the ball that Scott had hit? Write your answer in a complete sentence. 299

Problem 29 WORKED EXAMPLE Applications of Quadratic Inequalities A company s monthly revenue earned from selling!x items is given by the function ( ) = 680x and the monthly cost is given by C x! ( ) = 10000 2x 2. Write a function!! R x P(x) that represents the company s monthly profit from selling!x items then use that function to determine how many items the company must sell in order to make at least $100,000 in profits for a given month. To determine! P x ( ), compute as follows: ( ) = R( x) C( x) = 680x ( 10000 2x ) 2 = 680 10000 + 2x 2! P x Therefore, P x! ( ) = 2x 2 +680x 10000 which is a quadratic function modeling the company s monthly profit from selling x items. To determine how many items the company must sell in order to make at least $100,000 in profits, we must solve the inequality: P( x) 100,000 or! 2x 2 + 680x 10000 100,000 Trying to factor the quadratic would be difficult, if not impossible, therefore we will use a graphing approach. The functions, graphing window and graph are shown below. For the company to make at least $100,000 profit, they must sell 120 units or more each month. Notice that we rounded up to the nearest unit since selling 119.65 units would not make sense. However, if the intersection had been at 119.12 units, for example, we would still have rounded up to 120 units as selling 119 units would not make profit positive. 300

Problem 30 YOU TRY Applications of Quadratic Inequalities The model, B t! ( ) = 5t 2 20t +20759, gives the annual consumption of beef, in tons, in the U.S. where t represents the number of years since 2000. a) Use this model to determine in what year beef consumption was at a minimum. You will need to use a graphical solution process. Show your graph and work in the space below. Write your answer in a complete sentence. b) Use this model to predict the year in which beef consumption exceeds 25000 tons. You will need to use a graphical solution process. Show your graph and work in the space below. Round your answer to the nearest whole year. Write your answer in a complete sentence. 301