1. © 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 6 Algebra: Equations and Inequalities

2. © 2010 Pearson Prentice Hall. All rights reserved. 2 6.5 Quadratic Equations

3. © 2010 Pearson Prentice Hall. All rights reserved. 3 Objectives 1.Multiply binomials using the FOIL method. 2.Factor trinomials. 3.Solve quadratic equations by factoring. 4.Solve quadratic equations using the quadratic formula. 5.Solve problems modeled by quadratic equations.

4. © 2010 Pearson Prentice Hall. All rights reserved. 4 Multiplying Two Binomials Using the FOIL Method Binomial: An algebraic expression containing two terms in which each exponent that appears on the variable is a whole number. Examples: x + 3, x + 4, 3x + 4, 5x – 3 FOIL Method

5. © 2010 Pearson Prentice Hall. All rights reserved. 5 Example 1: Using the FOIL Method Multiply ( x + 3) ( x + 4) Solution: F: First terms= x x = x² O: Outside terms= x 4 = 4x I: Inside terms= 3 x = 3x L: Last terms= 3 4 = 12 (x + 3)( x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 combine like terms

6. © 2010 Pearson Prentice Hall. All rights reserved. 6 Factoring Trinomials Whose Leading Coefficient is 1 Trinomial: A simplified algebraic expression that contains three terms in which all variables have whole number exponents. We can use the FOIL method to multiply two binomials to obtain a trinomial: Factored Form F O I L Trinomial Form (x + 3)( x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 Factoring an algebraic expressing containing the sum or difference of terms means finding an equivalent expression that is a product.

7. © 2010 Pearson Prentice Hall. All rights reserved. 7 A Strategy for Factoring x² + bx + c

8. © 2010 Pearson Prentice Hall. All rights reserved. 8 Example 2: Factoring a Trinomial in x² + bx + c Factor: x² + 6x + 8. Solution: Step 1: Enter x as the first term of each factor: x² + 6x + 8 = (x )(x ) Step 2: List all pairs of factors of the constant, 8. Step 3. Try various combinations of these factors. The correct factorization is the one in which the sum of the Outside and Inside products is equal to 6x. Factors of 88,14,2  8,  1  4,  2

9. © 2010 Pearson Prentice Hall. All rights reserved. 9 Example 2: continued Possible Factorizations of x² + 6x + 8 Sum of Outside and Inside Products (Should Equal 6x) (x + 8)(x + 1) x + 8x = 9x (x + 4)(x + 2) 4x + 2x = 6x (x – 8)(x – 1)– 8x – x = – 9x (x – 4)(x –2)–2x – 4x = –6x Thus, x² + 6x + 8 = (x + 4 )(x + 2 ).

10. © 2010 Pearson Prentice Hall. All rights reserved. 10 Example 3: Factoring a Trinomial Factor: 3x²  20x + 28 Solution Step 1: Find two First terms whose product is 3x². 3x²  20x + 28 = (3x )(x ) Step 2: List all pairs of factors of 28. Because the middle term,  20x, is negative, both factors must be negative. The negative factorizations are:  1(  28),  2(  14),  4(  7) Step 3: Try various combinations of these factors:

11. © 2010 Pearson Prentice Hall. All rights reserved. 11 Example 3: continued Possible Factorizations of 3x²  20x + 28 Sum of Outside and Inside Products (Should Equal  20x) (3x – 1)(x – 28)  84x – x =  85x (3x – 28)(x – 1)  3x – 28x =  31x (3x – 2)(x – 14)  42x  2x =  44x (3x – 14)(x – 2)  6x  14x =  20x (3x – 4)(x – 7)  21x – 4x =  25x (3x – 7)(x – 4)  12x  7x =  19x Thus, 3x²  20x + 28 = (3x  14 )(x  2). Step 4: Verify using the FOIL method: (3x  14)(x  2) = 3x²  6x  14x + 28 = 3x²  20x + 28.

12. © 2010 Pearson Prentice Hall. All rights reserved. 12 Solving Quadratic Equations by Factoring A quadratic equation in x is an equation that can be written in the form: ax² + bx + c = 0 where a,b, and c are real numbers, with a  0. Zero-Product Principle If the product of two factors is zero, then one (or both) of the factors must have a value of zero. If A·B = 0, then A = 0 or B = 0

13. © 2010 Pearson Prentice Hall. All rights reserved. 13 Solving a Quadratic Equation by Factoring 1.If necessary, rewrite the equation in the form ax² + bx + c = 0, moving all terms to one side, thereby obtaining zero on the other. 2.Factor. 3.Apply the zero-product principle, setting each factor equal to zero. 4.Solve the equations created in step 3. 5.Check the solutions in the original equation.

14. © 2010 Pearson Prentice Hall. All rights reserved. 14 Example 8: Solving a Quadratic Equation by Factoring Solve: x²  2x = 35 Step 1: Move all the terms to one side by subtracting 35 from both sides: x²  2x  35= 0 Step 2: Factor. (x – 7)(x + 5) = 0 Steps 3 and 4: Set each factor equal to zero and solve each resulting equation: x – 7 = 0 or x + 5 = 0 x = 7 x =  5

15. © 2010 Pearson Prentice Hall. All rights reserved. 15 Example 8: continued Step 5: Check the solutions in the original equation: x²  2x = 35 Check 7Check  5 7²  2· 7 = 35(  5) ²  2(  5) = 35 49  14 = 35 25 +10 = 35 35 = 35 35 = 35 Both solutions are true. If a solution is false in the original equation, it is said to be an extraneous solution.

16. © 2010 Pearson Prentice Hall. All rights reserved. 16 Solving Quadratic Equations Using the Quadratic Formula The solutions of the quadratic equation in the form ax² + bx + c = 0, with a  0, are given by the quadratic formula:

17. © 2010 Pearson Prentice Hall. All rights reserved. Quadratic Formula 17 1. 2. 3. 4. 5. 6.

18. © 2010 Pearson Prentice Hall. All rights reserved. Quadratic Formula 18 6. 7. 8. 9. 10. 11.

19. © 2010 Pearson Prentice Hall. All rights reserved. 19 Example 10: Solving a Quadratic Equation Using the Quadratic Formula Solve: 2x² + 9x – 5 = 0 Solution: a = 2, b = 9, c =  5

20. © 2010 Pearson Prentice Hall. All rights reserved. 20 Example 10: Solving a Quadratic Equation Using the Quadratic Formula Solve: 2x² + 9x – 5 = 0 Solution: a = 2, b = 9, c =  5

21. © 2010 Pearson Prentice Hall. All rights reserved. 21 Example 11: Solving a Quadratic Equation Using the Quadratic Formula Solve 2x² = 4x + 1 Solution: Move all the terms to the right: 2x²  4x – 1 = 0a = 2, b =  4 and c =  1 Use the quadratic formula: Note: If the expression under the square root simplifies to a negative number, then the quadratic equation has no real solutions.

22. © 2010 Pearson Prentice Hall. All rights reserved. Your Turn Solve the following quadratic equations using the quadratic formula. 1.3x 2 + 2x – 3 = 0 2.4x 2 – 3x = 7 3.5x 2 – 1x = – 4 22

23. © 2010 Pearson Prentice Hall. All rights reserved. 23 Example 12: Blood Pressure and Age A person's normal systolic blood pressure measured in millimeters of mercury depends on his or her age – for men, according to the formula: P = 0.006A²  0.02A + 120 Find the age, to the nearest year, of a man whose systolic blood pressure is 125 mm Hg.

24. © 2010 Pearson Prentice Hall. All rights reserved. 24 Example 12: continued Solution: From the formula, 125 = 0.006A² – 0.02A + 120. Rewriting the equation as a = 0.006A² – 0.02A – 5 = 0, we get the values a = 0.006, b =  0.02 and c =  5 Therefore, 31 is the approximate age for a man with a normal systolic blood pressure of 125 mm Hg.