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3 Objectives 1.Use the zero-factor property to solve a quadratic equation. 2.Use the square root property to solve a quadratic equation. 3.Solve an application involving a quadratic equation. 1 1 2 2 3 3

4 Use the zero-factor property to solve a quadratic equation 1.

5 Use the zero-factor property to solve a quadratic equation We know that the definition of a quadratic equation. A quadratic equation in one variable is an equation of the form ax 2 + bx + c = 0 (a  0) where a, b, and c are real numbers. We have solved quadratic equations by factoring and using the zero-factor property. If ab = 0, then a = 0 or b = 0.

6 Use the zero-factor property to solve a quadratic equation The zero-factor property states that when the product of two numbers is 0, at least one of the numbers is 0. For example, the equation (x – 4)(x + 5) = 0 indicates that a product is equal to 0. By the zero-factor property, one of the factors must be 0: x – 4 = 0orx + 5 = 0 We can solve each of these linear equations to get x = 4 or x = –5 The equation (x – 4)(x + 5) = 0 has two solutions: 4 and –5.

7 Use the zero-factor property to solve a quadratic equation Factoring Method To solve a quadratic equation by factoring, we 1. Write the equation in ax 2 + bx + c = 0 form (called quadratic or standard form). 2. Factor the left side of the equation. 3. Use the zero-factor property to set each factor equal to 0. 4. Solve each resulting linear equation. 5. Check each solution in the original equation.

8 Example Solve: 6x 2 – 3x = 0. Solution: Since the equation is already in quadratic form, we begin by factoring the left side of the equation. 6x 2 – 3x = 0 3x(2x – 1) = 0 By the zero-factor property, we have 3x = 0 or 2x – 1 = 0 Factor out 3x, the GCF.

9 Example – Solution We can solve these linear equations to get 3x = 0 or 2x – 1 = 0 x = 0 2x = 1 Check: cont’d

10 Example – Solution The solutions are 0 and. cont’d

11 Use the zero-factor property to solve a quadratic equation The factoring method doesn’t work with many quadratic equations. For example, the trinomial in the equation x 2 + 5x + 1 = 0 cannot be factored by using rational coefficients. To solve such equations, we need to develop other methods. The first of these methods uses the square root property.

12 Use the square root property to solve a quadratic equation 2.

13 Use the square root property to solve a quadratic equation If x 2 = c, then x is a number whose square is c. Since and, the equation x 2 = c has two solutions. Square-Root Property The equation x 2 = c has two solutions: or We can write the result with double-sign notation. The equation (read as “x equals plus or minus ”) means that or.

14 Example Solve: x 2 = 16. Solution: The equation x 2 = 16 has two solutions: or Using double-sign notation, we have.

15 Example – Solution Check: The solutions are 4 and –4. cont’d

16 Solve an application involving a quadratic equation 3.

17 Example – Integers The product of the first and third of three consecutive odd integers is 77. Find the integers. Analyze the problem Some examples of three consecutive odd integers are 3, 5, and 7 and –29, –27, and –25 These examples illustrate that to obtain the next consecutive odd integer from the previous one, we must add 2.

18 Example – Integer Problems If we let x represent the first of three consecutive odd integers, then x + 2 represents the second odd integer and x + 2 + 2 or x + 4 represents the third odd integer. Form an equation Since the product of the first integer and the third integer is x(x + 4) and we are given that this product is 77, we have the equation x(x + 4) = 77 cont’d

19 Example – Integer Problems Solve the equation We can solve the equation as follows: x(x + 4) = 77 x 2 + 4x – 77 = 0 (x + 11)(x – 7) = 0 x + 11 = 0 or x – 7 = 0 x = –11 x = 7 Apply the distributive property and subtract 77 from both sides. Factor. Set each factor equal to 0. cont’d

20 Example – Integer Problems State the conclusion Since the first consecutive odd integer x can be either 7 or –11, there are two possible sets of integers. If the first integer is 7, the second is 9, and the third is 11. If the first integer is –11, the second is –9, and the third is –7. Check the results If the three consecutive odd integers are 7, 9, and 11, the product of the first and third integers is 7(11) = 77. cont’d

21 Example – Integer Problems If the three consecutive odd integers are –11, –9, and –7, the product of the first and third integers is –11(–7) = 77. The answers check and the integers are 7, 9, and 11 or –11, –9, and –7. cont’d