Lesson 21 Not So Dramatic Quadratics

STUDENT MANUAL ALGEBRA II / LESSON 21 Lesson 21 Not So Dramatic Quadratics Quadratic equations are probably one of the most popular types of equations that you ll see in algebra. A quadratic equation has at least one term whose exponent is to the second power. There can be no greater power in a quadratic equation bet you re happy to leave those cube roots in the dust for now! The general form of a quadratic equation is as follows: ax 2 + bx + c = 0 Generally, terms such as a, b, and c are used to represent constants. In the upcoming lessons, you ll learn all about quadratic equations, and find ways to solve both simple and complex quadratic equations. First, let s look at solving quadratic equations with simple square-root solutions. Let s take the example equation x 2 = 16 We can use the general square-root rule that shows x = + 16 In the square-root rule, there are always two possible solutions: the variable can either be positive or negative, since the product of two negative numbers is always positive. So for the equation x 2 = 16 the solution can either be + 4 or 4. This solution and rule might seem about as easy as could be, but what if there were a coefficient in the equation in front of the variable, such as: 5x 2 = 125 This equation doesn t necessarily follow the square root rule. The solution itself is actually quite simple though because you can easily divide the coefficient into both side of the equation: 5x 2 125 = x 2 125 5 = 25 So now we have: x 2 = 25 And this makes it easy to use the square-root rule to find the solution by taking the square root of each side: x 2 = + 25 x = +5

STUDENT MANUAL ALGEBRA II / LESSON 21 It seemingly obvious that you would choose to use the square-root rule when you have an equation that consists of a squared variable with a perfectly squared number, but what about when the number isn t a perfect square? Ugh. Well, it s actually not that complex either. Let s say you have the following equation: x 2 = 90 All you have to do determine one factor of the number that is a perfect square; this is known as the law of radicals, which states that you can separate the quantity under the radical. So, for 90, we can easily take out the square of 9 and 10 because 9 10 = 90 Now, x 2 = + 90 Becomes x = + 9 10 Or more simply written: x = +3 10 So the solution to the original quadratic equation: x 2 = 90 is x = +3 10. That s simpler, right? Yes, you re pretty much done from here. Basically, there are no more perfect squares in 10, so it s best to just keep it the way it is. When a number isn t a perfect square, such as the case with 10, we refer to it as irrational because the decimal value would never end.

STUDENT MANUAL ALGEBRA II / LESSON 21 Lesson 21 Not So Dramatic Quadratics Name: Date: Now, on your own, use the square-root rule and the law of radicals to solve for the variable in each given equation. Show all your work. 1. y 2 = 144 2. 225 = x 2 3. m 2 = 81 4. 400 = t 2

STUDENT MANUAL ALGEBRA II / LESSON 21 5. 525 = y 2 6. 4a 2 = 64 7. 3x 2 = 90 8. 2y 2 = 8 9. 10x 2 = 100

STUDENT MANUAL ALGEBRA II / LESSON 21 10. 11a 2 = 99 Now, turn the following solutions into quadratic equations with no coefficients or radicals. Show all your work. 11. y = +2 11 12. a = +4 13. x = +3 3

STUDENT MANUAL ALGEBRA II / LESSON 21 14. s = +6 6 15. x = +9 16. n = + 3 17. x = +5 7 18. a = +12 11

STUDENT MANUAL ALGEBRA II / LESSON 21 19. b = +20 10 20. z = +15 6

STUDENT MANUAL ALGEBRA II / LESSON 22 Lesson 22 Zero s Got Some Principles! Remember from the previous lesson that a quadratic equation has at least one term whose exponent is to the second power. The general form of a quadratic equation is as follows: ax 2 + bx + c = 0 Oftentimes quadratic equations aren t written solely as trinomials though, they are also often written in binomial form. One thing that is always essential when solving for quadratic equations is that you set the equation equal to zero; this way, you can easily determine the possible values of the variable. For example, let s say we have: 2x 2 6x = 0 Our equation is already set to zero, but in order to figure out the possible values of x, we must first find the GCF or greatest common factor remember that from some previous lessons ago? The GCF of 2x 2 6x is obviously 2x, so now, let s factor that out of the original binomial so we can solve. Why do we factor x out of the equation? Well it s simply because it s always much easier to solve when we don t have to deal with radical expressions. So let s rewrite the equation factored, and make sure that we set it to zero: 2x(x-3) = 0 Now, we have two obvious solutions for x. It must either be 0 or +3. How do we know this? Well, of x were to equal zero, then the equation would be true, let s take a look: 2(0)(0-3) 2 0 = 0 We distribute the 0 to the -3 and multiply to get 0 So the equation becomes 0=0, which is true! Let s see how it works if the solution for x is +3 2(3)(3-3) 2 3 = 6 3 3= 0 Any number times zero equals zero, right?

STUDENT MANUAL ALGEBRA II / LESSON 22 6 0 = 0 So the equation becomes 0=0, which is also true! So in this equation x has two solutions, it can either be 0 or 3, so we can write the answer this way: x = 0, 3 This denotes that x has two possible values. Either of which would make the equation true. The Zero Factor Principle states that in order to make any number become zero, you just simply multiply it by zero! Simple enough, right? So how can we apply this rule to quadratic equations? It s easy! Let s say we have the following equation: (x 4)(x +3) = 0 Well, based on the zero factor principle, I know that if one of the binomials is equal to zero, then the equation is true because the product of any number and zero equals zero. So we know that one of the two binomials that are being multiplied in the equation must be equal to zero in order for this statement to be true. Let s look at each binomial in the equation separately, and set them each to zero to determine the possible value of x by solving for x: (x 4)(x +3) = 0 x 4= 0 In order to find the value for x, we have to get it by itself on one side of the equation. In order to do this, we have to add 4 to both sides of the equation: x 4= 0

STUDENT MANUAL ALGEBRA II / LESSON 22 +4 +4 The -4 cancels out on the left side, and out solution becomes: x = 4 Let s see if x = 4 makes the original equation a true statement: (x 4)(x +3) = 0 (4 4)(4 +3) = 0 (0)(7) = 0 This is true! 0 7 = 0 0=0 So, one possible solution to x is 4 But wait! There is also another possible solution. What if the other binomial (x + 3) was equal to 0? Well let s set it equal to zero and see what the other possible solution could be: x + 3= 0

STUDENT MANUAL ALGEBRA II / LESSON 22 In order to find the value for x, we have to get it by itself on one side of the equation. In order to do this, we have to subtract 3 from both sides of the equation: x + 3= 0-3 -3 The + 3 cancels out on the left side, and out solution becomes: x = - 3 Let s see if x = -3 makes the original equation a true statement: (x 4)(-3 +3) = 0 (- 3 4)(- 3 +3) = 0 (-7)(0) = 0 This is true! -7 0 = 0 0=0 So, another possible solution to x is -3 Going through the process of setting each individual binomial in the equation to zero is known as checking. We checked to make sure that the statements were true.

STUDENT MANUAL ALGEBRA II / LESSON 22 Based on the zero factor principle, when the equation is set to zero, we can determine that there are two possible solutions for the equation: (x 4)(x +3) = 0 x = 4, -3 Remember, if there is a coefficient in front of the variable, then you must divide in order to solve. For example, if you have 2x 1 = 0 Then you must first add +1 to each side: 2x 1 = 0 +1 +1 2x=1 And next, you must get the x by itself by doing the opposite of what is being done to it to get it alone. So in this equation x is being multiplied by 2. What is the opposite of multiplication? That s right division! So now we need to divide both sides by 2 (remember all this stuff from Algebra 1?) 2x=1 2 2

STUDENT MANUAL ALGEBRA II / LESSON 22 The 2 cancels out on the left side, and we are left with x = ½ Let s do a check to see if this is true! Replace the x value with ½ in the original equation to see if it is true. 2x 1 = 0 2(½) 1 = 0 2 1 = 2 1 2 2 2 2 = 1 So, let s plug 1 into the original equation: 1 1 = 0 0 = 0 Yep, that s true! So we know that in the original equation : 2x 1 = 0 x= ½

STUDENT MANUAL ALGEBRA II / LESSON 22 Lesson 22 Zero s Got Some Principles! Name: Date: Now, on your own, find the greatest common factor in each equation and factor them out to determine all possible solutions for each given variable. Do a check and plug each of your possible solutions for the given variable in the equation to see if it makes the equation a true statement. Show all your work. 1. 3x 2 18x = 0 2. 4y 2 16y = 0 3. x 2 + 10x = 0

STUDENT MANUAL ALGEBRA II / LESSON 22 4. 5a 2 25a = 0 5. b 2 + 1/3b = 0 6. 6m 2 24m = 0

STUDENT MANUAL ALGEBRA II / LESSON 22 7. 12c 2 + 36c = 0 8. 8p 2 72p = 0 9. ½ x 2 + 4x = 0 10. ¾ y 2 1 ½ y = 0

STUDENT MANUAL ALGEBRA II / LESSON 22 Now, using the zero factor principle to determine all possible solutions for each given variable. Put all your answers into the simplest form. Do a check and plug each of your possible solutions for the given variable in the equation to see if it makes the equation a true statement. Show all your work. 11. (x 5)(x +6) = 0 12. (y 6)(y -6) = 0 13. (a 1)(a +1) = 0 14. (3y 1)(x +3) = 0

STUDENT MANUAL ALGEBRA II / LESSON 22 15. (6n + 9)(n +3/2) = 0 16. (2p + 4)(4p +2) = 0 17. (5x + 20)(1/2x +10) = 0

STUDENT MANUAL ALGEBRA II / LESSON 22 18. (3/2 b + 1)(8x +64) = 0 For the following equations, determine and prove if the solutions of the given variables are true or false. Show all your work. 19. For the equation: (3/4 y - 1)(1/5y + 5) = 0 Do the solutions: y = -4/3, -25 make the equation true? 20. For the equation: (5/6x 6/5)(x + 2) = 0 Do the solutions: x = +1, -2 make the equation true?

STUDENT MANUAL ALGEBRA II / LESSON 23 Lesson 23 As a Matter of Factor In the previous lesson, we discussed how to solve for a given variable when an equation is set to zero. For example, we determined that for: (x + 4)(x +3) = 0 x =- 4, -3 Now, (x + 4)(x +3) is a factored form of a quadratic equation. In fact, it s the factor of: x 2 + 7x +12=0 How do we know that (x + 4)(x +3) = 0is the factored form of the following quadratic equation: x 2 + 7x +12=0? Well, first of all, we use the FOIL method to turn the two binomials: (x + 4)(x +3) = 0 into a quadratic equation. Remember FOIL? It stands for First, Inner, Outer, Last, using the distributive property. Let s take a look: (x + 4)(x +3) = 0 F F

STUDENT MANUAL ALGEBRA II / LESSON 23 First: x x = x 2 (x + 4)(x +3) = 0 I I Inner: x 3 = 3x (x + 4)(x +3) = 0 O O Outer: 4 x = 4x (x + 4)(x +3) = 0 L L Outer: 4 3 = 12 So we have the products of the distribution: F: x 2 I: 3x O: 4x L: 12

STUDENT MANUAL ALGEBRA II / LESSON 23 x 2 + 3x + 4x + 12 = 0 x 2 + 7x + 12 = 0 Alright, so it s pretty simple to solve for a variable, when a quadratic equation is already factored, but what about when it s not? What if the equation set to zero is still in the quadratic form, such as: x 2 + 10x + 24 = 0 How can we solve then for x? We must first factor out the trinomial form of the quadratic equation. Ugh, so how can we factor this quadratic equation? First, look at the last constant number in the equation: 24 Now, think of a few factors that make up 24 and list them: 2 12 = 24 3 8 = 24 4 6 = 24 Remember, those factors must add up to the middle term in the equation: 10 2 12 = 24/ 2 +12= 14

STUDENT MANUAL ALGEBRA II / LESSON 23 3 8 = 24/ 3 +8 = 11 4 6 = 24/ 4 + 6 = 10-BINGO! So now we can factor out the equation so that we can solve for x: (x + 6)(x - 4) = 0 x= -6, +4 Be sure to plug in each assumed value for x into the equation to check for correctness! Okay, this was all nice and fine and dandy, but what happens when the quadratic equation looks something like this: 2x 2 + 11x + 12 = 0 Now there s a coefficient in front of the leading term, so we have to consider how that changes things when we go about factoring. Well, let s write out the equation so far as we know it: (2x + ) (x + ) Now, let s list all the factors of 12: 1 12 = 12

STUDENT MANUAL ALGEBRA II / LESSON 23 2 6 = 12 3 4 = 12 Now, take the portion of the equation that we already know, and test all the optional factors to see which one works: 1 12 = 12/ (2x + 1) (x +12) = 0/ 2x 2 + 2x + 24x + 12 = 0/ = 2x 2 + 26x + 12 = 0 2 6 = 12/ (2x + 2) (x +6) = 0/ 2x 2 + 12x + 2x + 12 = 0/ = 2x 2 + 14x + 12 = 0 3 4 = 12/(2x + 3) (x + 4) = 0/ 2x 2 + 8x + 3x + 12 = 0/ = 2x 2 + 11x + 12 = 0 So from our testing, we ve determined that the factors of 3 and 4 work for the factorization of our equation: 2x 2 + 11x + 12 = 0 = (2x + 3) (x + 4) = 0 Now, we just simply need to solve for the variable x by setting each portion of the equation to zero. 2x +3 = 0 2 2 x + 3/2 = 0

STUDENT MANUAL ALGEBRA II / LESSON 23-3/2 x + 4 = 0 x = -3/2-4 -4 x = -4 So the solution to the original equation of 2x 2 + 11x + 12 = 0 Is that x= -3/2, -4

STUDENT MANUAL ALGEBRA II / LESSON 23 Lesson 23 As a Matter of Factor Name: Date: Now, on your own, factor the following quadratic equations and determine all possible solutions for each given variable. Be sure to identify the factors of the equation and the possible solutions. Do a check and use the foil method to double-check your factorization. Show all your work. 1. x 2 + 3x + 2 = 0 2. y 2 + 18y + 80 = 0 3. a 2 4a - 5 = 0

STUDENT MANUAL ALGEBRA II / LESSON 23 4. x 2 5x - 24 = 0 5. y 2 6y - 40 = 0 6. a 2 11a + 30 = 0 7. p 2 9p + 8 = 0 8. y 2 + 14y + 48 = 0

STUDENT MANUAL ALGEBRA II / LESSON 23 9. a 2 + 17a + 72 = 0 10. x 2-12x - 45 = 0 11. 4x 2-14x + 6 = 0 12. 3p 2 11p - 20 = 0

STUDENT MANUAL ALGEBRA II / LESSON 23 13. 3y 2 + 5y - 2 = 0 14. 3a 2 + 22a + 24 = 0 15. 5x 2 + 24x + 16 = 0 16. 5x 2 + 47x + 18 = 0

STUDENT MANUAL ALGEBRA II / LESSON 23 17. 3y 2 + 28y + 49 = 0 18. 7a 2 + 29a + 4 = 0 19. 8x 2 + 6x - 2 = 0 20. 6p 2 4p - 10 = 0

STUDENT MANUAL ALGEBRA II / LESSON 24 Lesson 24 Don t Forget to Set it! In the previous lesson, we discussed how to solve for a given variable in a quadratic equation that has not yet been factored. Well, there are still a few situations that you might face when running into solving for a variable in a quadratic equation. Let s look at the following equation: 7x + x 2 = -12 Yikes! Do you see what s wrong with the above problem? There s actually a couple problems with this equation, or really just a few little things that we need to fix before we can go about solving it. First of all, we need to make sure that the equation is set to equal zero. So what do we need to do to make the equation 7x + x 2 = - 12 equal to zero? Well, this one is pretty simple, we just add 12 to both sides: 7x + x 2 = -12 +12 +12 Now, the equation becomes: 7x + x 2 + 12 = 0 Wait, we still have a problem Houston! This equation isn t in the correct form.

STUDENT MANUAL ALGEBRA II / LESSON 24 The general quadratic formula should be in the following form: ax 2 + bx + c = 0 In that standard from a= the coefficient of the squared term, b is the coefficient of the firstdegree term, and c represents the constant. We ll work with this form a bit more later on. Okay, so let s rearrange our equation so that it s in proper form: 7x + x 2 + 12 = 0 x 2 + 7x + 12 = 0 Great! Now, we can factor the quadratic equation and solve for the variable x: (x + 4)(x +3) = 0 So now we know that x =- 4, -3

STUDENT MANUAL ALGEBRA II / LESSON 24 Lesson 24 Don t Forget to Set it! Name: Date: Now, on your own, find the solutions for each variable given. Show all your work. In order to do this, remember, you must follow this process: 1. Set each equation to be equal to zero. 2. Then put the equation in the quadratic ax 2 + bx + c = 0 form. 3. Factor the following quadratic equations. 4. Determine all possible solutions for each given variable. 1. 3x + 2 = - x 2 2. y 2-40 = + 6y 3. 3s + 1 = -2s 2

STUDENT MANUAL ALGEBRA II / LESSON 24 4. - 12 + x= - x 2 5. 2t 2-20 = + 3t 6. 3s + 2 = -s 2

STUDENT MANUAL ALGEBRA II / LESSON 24 7. y 2 + 18y = -80 8. p - 6 = -p 2 9. x 2-5 = 4x 10. t 2-24 = + 5t

STUDENT MANUAL ALGEBRA II / LESSON 24 11. 7g 2 + 4 = - 29g 12. 12s - 14 = -2s 2 13. 3y 2-28y = -49 14. 47z + z 2 = - 18

STUDENT MANUAL ALGEBRA II / LESSON 24 15. 15 + x 2 = - 13x 16. 2g 2 + 20 = - 14g 17. 5s - 2 = -3s 2 18. 3y 2 + 22y = -24

STUDENT MANUAL ALGEBRA II / LESSON 24 19. 24z + 5z 2 = - 16 20. 12 + 5x 2 = 19x

STUDENT MANUAL ALGEBRA II / LESSON 25 Lesson 25 Back at it with Quadratics As you may have noticed from the previous lessons, there are many different issues that you might run into when trying to solve for a given variable in a quadratic equation. Well, we re going to go over just one more problem that might arise when trying to solve for a given variable in a quadratic equation; although, there are several other problems that might arise, but we ll just going to cover the most common. Okay, so let s say that you re given the following equation: (x + 2) (x + 5) = - 2 Hey, this looks great, looks like it s already factored for us, right? So this should be easy enough Well, we ve got another problem; this equation isn t set to zero! Oh no! So now, we ve got to use the foil method so that we can set the equation to zero. So when we foil out: (x + 2) (x + 5) = - 2 We get: x 2 + 7x +10= - 2 So we can easily set it to zero now, by simply adding 2 to each side of the equation. x 2 + 7x +10= - 2 +2 +2

STUDENT MANUAL ALGEBRA II / LESSON 25 Now the equation is finally set to zero, so we have: x 2 + 7x +12=0 So remember at the very beginning of this all when we thought we could get away without factoring our quadratic equation? Well, now we re going to have to start all over again and factor our new equation: x 2 + 7x +12=0 And we get: (x + 4)(x +3) = 0 Now, we can solve and determine that: x = -4, -5

STUDENT MANUAL ALGEBRA II / LESSON 25 Lesson 25 Back at it with Quadratics Name: Date: Now, on your own, distribute to solve the following equations, and then set the quadratic equation to zero. Then, factor the new quadratic equation and solve. Reduce all solutions as much as possible, but keep your answers in rational form. Be sure to include all your steps for finding the solutions. 1. (x + 2) (x + 1) = 1 2. (y + 10) (y + 8) = 3 3. (a + 5)(a + 4) =6

STUDENT MANUAL ALGEBRA II / LESSON 25 4. (x - 3)(x - 5) =6 5. (y - 10) (y + 4) = -45 6. (a - 5) (a - 6) =19 7. (p - 1) (p - 8) =10

STUDENT MANUAL ALGEBRA II / LESSON 25 8. (y + 8) (y + 6) =8 9. (a + 8) (a + 9) =6 10. (x - 15) (x + 3)=77

STUDENT MANUAL ALGEBRA II / LESSON 25 11. (4x - 2) (x - 3) =14 12. (3p + 4) (p - 5) =22 13. (3y - 1) (y + 2)=26 14. (3y - 1) (y + 2) =76

STUDENT MANUAL ALGEBRA II / LESSON 25 15. (3a + 4) (a + 6) =17 16. (5x + 4) (x + 4) =276 17. (5x + 2) (x + 9) =-38 18. =(3y - 7) (y - 7) =-15

STUDENT MANUAL ALGEBRA II / LESSON 25 19. =(7a + 1) (a + 4) =-20 20. (4x -1) (2x + 2) =-25

STUDENT MANUAL ALGEBRA II / LESSON 26 Lesson 26 Finally the Formula! Okay, so we ve come across a ton of ways that the quadratic formula can mess us up when we re trying to solve for a given variable. Now, we re going to look at one pretty complex, but somewhat sure-fire way to solve quadratic equations. We re going to learn about using the quadratic formula when we have a quadratic equation that just won t factor. When there are such un-factorable equations, that do have solutions, then their answers are generally presented with irrational numbers in their answers. Irrational numbers are numbers that don t really have fractional equivalents. Irrational numbers have decimal values that just keep going and going and going! The quadratic formula can be used to solve equations with rational solutions, but generally, it s just easiest to use the factoring techniques that we discussed in the previous lessons for those type of quadratic equations. Now, we re going to focus on using the quadratic formula to help us solve some of those equations with irrational solutions. The quadratic formula states that when you have a quadratic equation in the following form: ax 2 + bx + c = 0 With a as the coefficient of the leading term, b as the coefficient of the first-degree term, and c as the constant. We can solve the equation by using the following solution: x = -b ± b 2 4ac 2a Looks fun, huh? Ah, it s not so bad. We can do this, no problem. But before you re given an equation to solve using the quadratic formula, be aware of the following things you must keep in mind as you solve: -Remember to find the opposite of b

STUDENT MANUAL ALGEBRA II / LESSON 26 -Remember that you must solve for both the positive and negative possible solutions under the radical sign. -Remember to correctly simplifying the numbers that are under the radical sign. -Remember to divide the whole equation by the denominator. Now, let s try to solve the following equation using the quadratic formula: 2x 2 + 5x -6= 0 Let: a = 2 b = 5 c = -6 Now, let s plug it in: x = -b ± b 2 4ac 2a x = -5 ± 25 + 48 4 x = -5 ± 73 4

STUDENT MANUAL ALGEBRA II / LESSON 26 Plug those numbers into a trusty calculator, and: x = -5 + 73 = approximately: 0.886 4 x = -5-73 = approximately: -3.386 4 If you come up with a solution where there is a negative number under the radical sign, then you should know that the problem has no real solution. That s basically it with the quadratic formula!

STUDENT MANUAL ALGEBRA II / LESSON 26 Lesson 26 Finally the Formula! Name: Date: Identify the values of a, b, and c in the following equations for the quadratic formula. 1. y 2 + 14y + 48 = 0 2. x 2-12x - 45 = 0 3. 4x 2-14x + 6 = 0

STUDENT MANUAL ALGEBRA II / LESSON 26 4. 3p 2 11p - 20 = 0 5. 3y 2 + 5y - 2 = 0 Place the values of a, b, and c of the following equations into the quadratic formula. Do not solve. 6. 3x 2 + 22x + 24 = 0 7. 5x 2 + 24x + 16 = 0

STUDENT MANUAL ALGEBRA II / LESSON 26 8. 5x 2 + 47x + 18 = 0 9. 3y 2-28y + 49 = 0 10. 7x 2 29x + 4 = 0

STUDENT MANUAL ALGEBRA II / LESSON 26 Now, on your own, use the quadratic formula to solve all of the equations below. Be sure to include all your steps in the equation for finding the solutions. Use a calculator to find the decimal value of your answer, approximate each solution two decimal places. Remember, you must first put the equation in ax 2 + bx + c = 0 form before you can solve. Show all your work. 11. 5x 2 + 24x + 16 = 0 12. 9x 2-30x + 25 = 0 13. 8x 2 + 18x - 5 = 0 14. x 2-5x + 2 = 0

STUDENT MANUAL ALGEBRA II / LESSON 26 15. 2x 2 + x - 4 = 0 16. x 2 + 1 = 3x 17. x 2 + 8 = 6x + 2 18. 2(x 2 x) = 5

STUDENT MANUAL ALGEBRA II / LESSON 26 19. 3(x 2-2x) = 5 2x See if you can recall the quadratic formula without looking back at your notes. 20.

STUDENT MANUAL ALGEBRA II / LESSON 28 Lesson 28 Cumulative Review for Lessons 21-27 Name: Date: Please complete the following cumulative review. Ask you instructor if you have any questions. You may want to review some of the instructional sections of previous lessons to help you remember certain processes as you complete the work below. These will be the same concepts covered in the upcoming assessment. Use the square-root rule and the law of radicals to solve for the variable in each given equation. Show all your work. 1. 525 = y 2 2. 4a 2 = 64 3. 3x 2 = 90

STUDENT MANUAL ALGEBRA II / LESSON 28 Turn the following solutions into quadratic equations with no coefficients or radicals. Show all your work. 4. x = +3 3 5. s = +6 6 Find the greatest common factor in each equation and factor them out to determine all possible solutions for each given variable. Do a check and plug each of your possible solutions for the given variable in the equation to see if it makes the equation a true statement. Show all your work. 6. 5a 2 25a = 0 7. b 2 + 1/3b = 0

STUDENT MANUAL ALGEBRA II / LESSON 28 Use the zero factor principle to determine all possible solutions for each given variable. Put all your answers into the simplest form. Do a check and plug each of your possible solutions for the given variable in the equation to see if it makes the equation a true statement. Show all your work. 8. (a 1)(a +1) = 0 9. (3y 1)(x +3) = 0 10. (6n + 9)(n +3/2) = 0

STUDENT MANUAL ALGEBRA II / LESSON 28 Factor the following quadratic equations and determine all possible solutions for each given variable. Be sure to identify the factors of the equation and the possible solutions. Do a check and use the foil method to double-check your factorization. Show all your work. 11. y 2 + 14y + 48 = 0 12. a 2 + 17a + 72 = 0 13. x 2-12x - 45 = 0

STUDENT MANUAL ALGEBRA II / LESSON 28 14. 4x 2-14x + 6 = 0 15. 3p 2 11p - 20 = 0 Find the solutions for each variable given, using factoring methods. Show all your work. 16. p - 6 = -p 2 17. x 2-5 = 4x

STUDENT MANUAL ALGEBRA II / LESSON 28 18. t 2-24 = + 5t 19. 7g 2 + 4 = - 29g 20. 12s - 14 = -2s 2

STUDENT MANUAL ALGEBRA II / LESSON 28 Distribute to solve the following equations, and then set the quadratic equation to zero. Then, factor the new quadratic equation and solve. Reduce all solutions as much as possible, but keep your answers in rational form. 21. (y + 8) (y + 6) =8 22. (a + 8) (a + 9) =6

STUDENT MANUAL ALGEBRA II / LESSON 28 23. (x - 15) (x + 3)=77 24. (4x - 2) (x - 3) =14 25. (3p + 4) (p - 5) =22 Be able to write the quadratic formula without looking at your notes 26.

STUDENT MANUAL ALGEBRA II / LESSON 28 Use the quadratic formula to solve all of the equations below. Be sure to include all your steps in the equation for finding the solutions. Use a calculator to find the decimal value of your answer, approximate each solution two decimal places. 27. 3(x 2-2x) = 5 2x 28. 9x 2-30x + 25 = 0 29. 8x 2 + 18x - 5 = 0 30. x 2-5x + 2 = 0