Section 7.1 Quadratic Equations

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1 Section 7.1 Quadratic Equations INTRODUCTION In Chapter 2 you learned about solving linear equations. In each of those, the highest power of any variable was 1. We will now take a look at solving quadratic equations, in which the highest variable power is 2, as in 3x 2 + 2x = 8. A quadratic equation is an equation in which the degree of the polynomial is two. A standard form quadratic equation looks like this: ax 2 + bx + c = 0. Notice that the left side of this equation is a polynomial and is in descending order and the right side is 0. Some quadratic eqautions do not start out being in standard form but can be rewritten to become standard form. Here are more examples of quadratic equations: - 2x 2 + 6x = 0 x 2 7x = - 10 x 2 = - 2x + 8 4x 2 9x + 3 = x 2 3 5x 4 = 2x 2 9x = 6x 2 13x 5 PREPARATION Before jumping in to solving quadratic equations, let s take some time to look at some introductory ideas. The first is understanding, how to quickly solve a special group of linear equations. This special group of linear equations are those of the form ax + b = 0. Here are some examples: 4x + 8 = 0 3x 12 = 0-2x 10 = 0 3x = 0 As always, we need to solve for x by isolating the variable. This time, however, we are going to solve each equation in just one step. In Section 3.1 we learned about moving a term from one side of the equation to the other side; when a whole term moves from one side of the equation to the other, it shows up again as its opposite: 3x + 12 = 0 We can move the constant term, 12, to the other side as x = - 12 Dividing each side by 3 leaves just x on the left side. x = We have isolated the variable. x = - 4 This is the solution; it makes the equation true. We can check out the solution by replacing the x with - 4: 3(- 4) + 12 = = 0 this is true, so the solution is x = - 4. Quadratic Equations page 7.1-1

2 In this section we want to be able to solve these special linear equations in just one step. To do so requires some mental calculations; these calculations are not difficult, but they do require some practice. Example 1: Solve each of these linear equations in just one step by isolating the variable. a) x + 5 = 0 b) x 3 = 0 c) x = 0 Procedure: For each, mentally move the constant term to the other side. As it moves to the other side, it shows up as its opposite. a) x + 5 = 0 Move + 5 to the other side: we get x = - 5 b) x 3 = 0 Move - 3 to the other side: we get x = 3 c) x = 0 Here, there is nothing to move; in fact, the variable is already isolated: x = 0! Exercise 1 Solve each of these linear equations in just one step by isolating the variable. a) x + 5 = 0 b) x 4 = 0 c) x 7 = 0 d) x = 0 e) x + 1 = 0 f) x 9 = 0 g) x 6 = 0 h) x + 3 = 0 i) x = 0 j) x + 9 = 0 k) x 12 = 0 l) x + 8 = 0 Quadratic Equations page 7.1-2

3 Example 2: Solve each of these linear equations in just one step by isolating the variable. Simplify whenever possible. a) 2x 10 = 0 b) 4x + 8 = 0 c) 2x 5 = 0 d) 3x + 7 = 0 e) 4x = 0 f) - 6x = 0 Procedure: For each, mentally move the constant term to the other side; then divide by the coefficient of x. a) Two steps: 2x 10 = 0 One step: 2x 10 = 0 Move the constant over: 2x = 10 x = 5 Divide by the coefficient: x = 10 2 = 5 b) Two steps: 4x + 8 = 0 One step: 4x + 8 = 0 Move the constant over: 4x = - 8 x = - 2 Divide by the coefficient: x = = - 2 c) Two steps: 2x 5 = 0 One step: 2x 5 = 0 Move the constant over: 2x = 5 x = 5 2 Divide by the coefficient: x = 5 2 d) Two steps: 3x + 7 = 0 One step: 3x + 7 = 0 Move the constant over: 3x = - 7 x = Divide by the coefficient: x = e) One step: 4x = 0 One step: 4x = 0 Divide by the coefficient: x = 0 4 = 0 x = 0 f) One step: - 6x = 0 One step: - 6x = 0 Divide by the coefficient: x = 0-6 = 0 x = 0 Quadratic Equations page 7.1-3

4 Exercise 2 Solve each of these linear equations in just one step. Simplify whenever possible. a) 2x + 6 = 0 b) 2x 10 = 0 c) 2x 7 = 0 d) 3x + 12 = 0 e) 3x 3 = 0 f) 3x 4 = 0 g) 4x + 7 = 0 h) 5x 2 = 0 i) 3x + 1 = 0 j) 2x = 0 k) 5x = 0 l) - 3x = 0 m) - x = 0 n) 4x 9 = 0 o) 6x + 4 = 0 p) - 4x = 0 q) - 2x = 0 r) 9x 9 = 0 Quadratic Equations page 7.1-4

5 THE ZERO PRODUCT PRINCIPLE You know that the product of two positive numbers is another positive number: (+ 2) (+ 3) = + 6, and that the product of two negative numbers is a positive number: (- 4) (- 5) = + 20, and that the product of a positive number and a negative number is a negative number: (+ 3) (- 7) = - 21, and (- 5) (+ 6) = - 30, It seems, by the examples above, that product of two numbers is either positive or negative. Is it ever possible that the product of two numbers is 0? Yes, but only if one, or both, of the numbers is zero. This is the idea behind the Zero Product Principle. The Zero Product Principle If the product of two numbers is 0, then one of the numbers must be 0. If A B = 0, then either A = 0 or B = 0. Example 3: What can be said about the values of A and B? a) A 4 = 0 b) 6 B = 0 c) A B = 0 Procedure: Apply the Zero product Principle to each equation. a) If A 4 = 0, then A must equal 0: A = 0. b) If 6 B = 0, then B must equal 0: B = 0. c) If A B = 0, then either A must be 0 or B must be 0: A = 0 or B = 0. Exercise 3 What can be said about the values of P and Q? a) P 2 = 0 b) 7 Q = 0 c) P Q = 0 Quadratic Equations page 7.1-5

6 PUTTING THE ZERO PRODUCT PRINCIPLE INTO PRACTICE The Zero Product Principle can be applied to any two factors that multiply to get 0. For example, if This is better written without the parentheses, as (x + 3)(x 4) = 0, then one of the factors must be 0: either (x + 3) = 0 or (x 4) = 0 either x + 3 = 0 or x 4 = 0 Notice that each of these is a linear equation. In fact, each of these is one of the special linear equations seen in Example 1, so they can easily be solved in one step: if x + 3 = 0 or x 4 = 0 then x = - 3 or x = 4 We could write this answer, the two solutions, as x = - 3, 4. Example 4: Use the Zero product Principle to solve each equation. a) (x + 6)(x 1) = 0 b) x(x + 5) = 0 c) (3x + 6)(2x 5) = 0 d) - 4x(6x + 7) = 0 Procedure: Each of these will separate into two linear equations that can be solved in one step. a) (x + 6)(x 1) = 0 b) x(x + 5) = 0 x + 6 = 0 or x 1 = 0 x = 0 or x + 5 = 0 x = - 6 or x = 1 x = 0 or x = - 5 x = - 6, 1 x = 0, c) (3x + 6)(2x 5) = 0 d) - 4x(6x + 7) = 0 3x + 6 = 0 or 2x 5 = 0-4x = 0 or 6x + 7 = 0 x = - 2 or x = 5 2 x = 0 or x = x = - 2, 5 2 x = 0, Quadratic Equations page 7.1-6

7 Exercise 4 Use the Zero product Principle to solve each equation. a) (x + 5)(x + 8) = 0 b) (x 3)(x 2) = 0 c) (2x + 5)(2x + 8) = 0 d) (3x 3)(3x 2) = 0 e) x(x + 1) = 0 f) 5x(6x 1) = 0 g) (x + 7)(x 7) = 0 h) (4x 5)(4x + 5) = 0 i) - 2x(3x 4) = 0 j) 12x(5x + 6) = 0 Quadratic Equations page 7.1-7

8 SOLVING QUADRATIC EQUATIONS BY FACTORING All of the examples and exercises you ve seen so far have been set up so that the polynomial is in factored form. If an equation is a quadratic equation, then it needs to be in factored form before we can solve it. For example, x 2 + 8x + 12 = 0 has the form of a typical quadratic equation. However, before we can apply the Zero Product Principle, we must factor the trinomial. We will then have the product of two factors equal to 0: x 2 + 8x + 12 = 0 Factor the polynomial: (x + 6)(x + 2) = 0 Write the two linear equations: x + 6 = 0 or x + 2 = 0 Solve these equations in one step: x = - 6 or x = - 2 Write the two solutions: x = - 6, - 2 It might be hard to believe that the numbers - 6 and - 2 are solutions to the equation. We can verify that they are solutions by doing a check. To do a check on a quadratic equation, we choose only one solution at a time. Every x in the equation is then replaced by that value; of course, we re looking for a true statement from each solution. Check x = - 6: does (- 6) 2 + 8(- 6) + 12 = 0? = = 0 This is true, so x = - 6 is a solution. Now check x = - 2: does (- 2) 2 + 8(- 2) + 12 = 0? = = 0 This is true, so x = - 2 is a solution. Since both of our answers led to true statements, we know that the solutions are x = - 6, - 2. Quadratic Equations page 7.1-8

9 Example 5: Factor the polynomial and use the Zero product Principle to solve each equation. a) x 2 + 2x 35 = 0 b) - 2x 2 8x = 0 c) 4x 2 13x + 3 = 0 d) x 2 25 = 0 Procedure: You should check the factoring for these So that you are confident about the results. a) x 2 + 2x 35 = 0 b) - 2x 2 8x = 0 (x + 7)(x 5) = 0-2x(x + 4) = 0 x + 7 = 0 or x 5 = 0-2x = 0 or x + 4 = 0 x = - 7 or x = 5 x = 0 or x = - 4 x = - 7, 5 x = 0, c) 4x 2 13x + 3 = 0 d) x 2 25 = 0 (4x 1)(x 3) = 0 (x + 5)(x 5) = 0 4x 1 = 0 or x 3 = 0 x + 5 = 0 or x 5 = 0 x = 1 4 or x = 3 x = - 5 or x = 5 x = 1 4, 3 x = - 5, 5 Exercise 5 Factor the polynomial and use the Zero product Principle to solve each equation. a) x 2 + 3x 28 = 0 b) 3x x = 0 Quadratic Equations page 7.1-9

10 Exercise 6 Factor the polynomial and use the Zero Product Principle to solve each equation. a) x 2 8x + 15 = 0 b) - 5x x = 0 c) x 2 4x 60 = 0 d) x 2 81 = 0 e) 4x 2 5x 6 = 0 f) 9x 2 16 = 0 Quadratic Equations page

11 WHY THERE IS NO TWELVE PRODUCT PRINCIPLE What if the quadratic equation is written so that the polynomial is not equal to 0? For example, in the equation x 2 + 3x + 2 = 12 can we factor the left side and set each factor equal to 12? Let s try it! x 2 + 3x + 2 = 12 Factor the polynomial: (x + 1)(x + 2) = 12 Write the two linear equations: x + 1 = 12 or x + 2 = 12 Solve these equations in one step: x = 11 or x = 10 To check these answers one at a time: Check x = 11: does (11) 2 + 3(11) + 2 = 12? = = 0 This is false! x = 11 is not a solution. If we were to check x = 10, we d find that it is also not a solution. Where was the error? The error is in thinking that all numbers behave as 0 does. If a product is equal to 12, as in (x + 1)(x + 2) = 12, then it s unpredictable what the value of each factor is. There are many options for the factors whose product is 12: 2 6, 3 4, - 1 (- 12), , and the list goes on and on This example is used to illustrate that there is only one product principle that applies to quadratic equations, the Zero Product Principle. In other words, to solve a quadratic equation, the polynomial must first be set equal to 0. Here s the proper way to solve x 2 + 3x + 2 = 12: x 2 + 3x + 2 = 12 First, subtract 12 from each side so that 0 is on one side: - 12 =- 12 Now factor the polynomial: x 2 + 3x 10 = 0 (x + 5)(x 2) = 0 Write the two linear equations: x + 5 = 0 or x 2 = 0 Solve these equations in one step: x = - 5 or x = 2 These are the solutions; you may check them for certainty: x = - 5, 2 Quadratic Equations page

12 Example 6: Solve each quadratic equation by first having one side become 0. a) x 2 5x + 2 = - 4 b) (x + 6)(x 3) = - 20 c) 4x x 2 = 4 Procedure: You should check the factoring for these so that you are confident about the results. a) Add 4 to each side to get 0 on the right side: x 2 5x + 2 = =+ 4 x 2 5x + 6 = 0 Now factor the polynomial: (x 3)(x 2) = 0 Set each linear factor equal to 0: x 3 = 0 or x 2 = 0 Solve each linear equation mentally: x = 3 or x = 2 Write the solutions: x = 3, 2 b) Since we need to get 0 on one side (the right side (x + 6)(x 3) = - 20 in this case),we ll need to first multiply the binomials so that we can see the polynomial: x 2 + 3x 18 = - 20 Add 20 to each side to get 0 on the right side: + 20 =+ 20 x 2 + 3x + 2 = 0 Now factor the polynomial: (x + 1)(x + 2) = 0 Set each linear factor equal to 0: x + 1 = 0 or x + 2 = 0 Solve each linear equation mentally: x = - 1 or x = - 2 Write the solutions: x = - 1, - 2 c) There are two unique aspects in this quadratic equation 4x x 2 = 4 First, the x 2 term is negative on the left side, so we should move the terms so that the polynomial is on the right side, in descending order, and 0 is on the left: 0 = x 2 4x + 4 This trinomial factors into two identical binomials: 0 = (x 2)(x 2) Set each linear factor equal to 0: x 2 = 0 or x 2 = 0 Solve each linear equation mentally: x = 2 or x = 2 Now the second unique aspect shows up; there s only one solution: x = 2 (only) Quadratic Equations page

13 Exercise 7 Solve each quadratic equation by first having one side become 0. Check your answers to verify that they are solutions. a) x 2 + 5x 8 = 16 b) 3x 2 + x = 10 c) (x + 4)(x 8) = - 20 d) x 2 + 2x = 8x 9 e) 6x = 5x 2 8 f) x 2 4x = 25 4x Quadratic Equations page

14 Answers to each Exercise Section 7.1 Exercise 1: a) x = - 5 b) x = 4 c) x = 7 d) x = 0 e) x = - 1 f) x = 9 g) x = 6 h) x = - 3 i) x = 0 j) x = - 9 k) x = 12 l) x = - 8 Exercise 2: a) x = - 3 b) x = 5 c) x = 7 2 d) x = - 4 e) x = 1 f) x = 4 3 g) x = h) x = 2 5 i) x = j) x = 0 k) x = 0 l) x = 0 m) x = 0 n) x = 9 4 o) x = p) x = 0 q) x = 0 r) x = 1 Exercise 3: a) P must equal 0; P = 0 b) Q must equal 0; Q = 0 c) either P must be 0 or Q must be 0; P = 0 or Q = 0 Exercise 4: a) x = - 5, - 8 b) x = 3, 2 c) x = - 5 2, - 4 d) x = 1, 2 3 e) x = 0, - 1 f) x = 0, 1 6 g) x = - 7, 7 h) x = 5 4, i) x = 0, 4 3 j) x = 0, Exercise 5: a) x = - 7, 4 b) x = 0, - 4 Exercise 6: a) x = 5, 3 b) x = 0, 4 c) x = 10, - 6 d) x = 9, - 9 e) x = - 3 4, 2 f) x = - 4 3, 4 3 Exercise 7: a) x = - 8, 3 b) x = 5 3 d) x = 3 e) x = - 4 5, - 2 c) x = 6, - 2, 2 f) x = 5, - 5 Quadratic Equations page

15 Section 7.1 Focus Exercises 1. Use the Zero Product Principle to solve each equation. a) (x + 3)(x 7) = 0 b) (x 2)(x 4) = 0 c) (4x 5)(2x + 9) = 0 d) (2x + 3)(5x 6) = 0 e) - 2x(x + 5) = 0 f) 3x(9x 5) = 0 g) (x + 8)(x 8) = 0 h) (2x 3)(2x + 3) = 0 2. Factor the polynomial and use the Zero Product Principle to solve each equation. a) x 2 25 = 0 b) x 2 49 = 0 c) 6x 2 54x = 0 d) - 3x x = 0 3. Factor the polynomial and use the Zero product Principle to solve each equation. Quadratic Equations page

16 a) x 2 9x + 20 = 0 b) x x + 35 = 0 c) x 2 x 90 = 0 d) x 2 + x 42 = 0 e) 4x 2 + 7x 15 = 0 f) 2x 2 13x + 15 = 0 g) 4x 2 9x 9 = 0 h) 6x x + 4 = 0 Quadratic Equations page

17 4. Solve each quadratic equation by first having one side become 0. Check your answers to verify that they are solutions. a) x 2 + 2x 4 = 59 b) x x 2 = 50x c) (x + 6)(x 7) = - 40 d) x 2 4x = 3x 10 e) x = 3x 2 10 f) x 2 x = 18 4x Quadratic Equations page