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Chapter 6 Equations and Inequalities

Chapter 6 Equations and Inequalities. 6.5 Quadratic Equations. Objectives. Learn the definition of quadratic equation. Multiply two binomials using the FOIL method. Factor trinomials. Solve quadratic equation by factoring. Solve quadratic equation using the quadratic formula

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Chapter 6 Equations and Inequalities

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  1. Chapter 6Equations and Inequalities

  2. 6.5Quadratic Equations

  3. Objectives • Learn the definition of quadratic equation. • Multiply two binomials using the FOIL method. • Factor trinomials. • Solve quadratic equation by factoring. • Solve quadratic equation using the quadratic formula • Use quadratic equations in applications.

  4. Multiplying Two Binomials Using FOIL Method • Binomial: has 2 terms • E.g., x + 3, 3x – 5 • FOIL Method

  5. Example • 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

  6. Factoring Trinomials • Trinomial: -- has 3 terms • Note:Factored Form F O I L Trinomial Form(x + 3)( x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 • Factoring an algebraic expression:--finding expression that is a product

  7. A Strategy for Factoring x² + bx + c

  8. Factoring a Trinomial in x² + bx + c (cont.) Factor: x² + 6x + 8. Solution: • Enter x as the first term of each factor: x² + 6x + 8 = (x )(x ) • List all pairs of Factors of constant, 8. • 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.

  9. Factoring a Trinomial in x² + bx + c (cont.) Thus, x² + 6x + 8 = (x + 4 )(x + 2 ).

  10. Example • Factor: 3x²  20x + 28 • Solution: • Factor: 3x²  20x + 28 3x²  20x + 28 = (3x )(x) • List all pairs of factors of 28. Because the middle term , 20x, is negative, both factors must be negative. The negative factorizations are: z1(28), 2(14), 4(7) • Try various combinations of these factors:

  11. Example (cont.) 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. Solving a Quadratic Equation by Factoring • Rewrite (in necessary) the equation in the formax2 + bx + c = 0. • Factor. • Let each factor equal to 0. • Solve the equations in step 3. • Check the solutions in the original equation.

  13. Example • Solve: x²  2x = 35 • Solution: • x²  2x35= 0 • (x – 7)(x + 5) = 0 • Set each factor equal to zero.x– 7 = 0 or x+ 5 = 0 • Solve each equation.x = 7 x = 5 Check:7²  2· 7 = 35 (5) ² 2(5) = 35 49 14 = 35 25 +10 = 35 35 = 35 35 = 35

  14. Quadratic Formula • Given:ax2 + bx + c = 0 • Solve for x.

  15. Quadratic Formula (cont.)

  16. Quadratic Formula • Given: ax2 + bx + c = 0, with a ≠ 0 • Solution is given by the Quadratic Formula:

  17. Example Solve: 2x² + 9x – 5 = 0 Solution: a = 2, b = 9, c = 5

  18. Example • Solve 2x² = 4x + 1 • Solution:2x² 4x – 1 = 0 a= 2, b = 4 and c = 1

  19. Your Turn • Solve the following quadratic equations using the quadratic formula. • 3x2 + 2x – 3 = 0 • 4x2 – 3x = 7 • 5x2 – 1x = – 4

  20. Your Turn • Use the quadratic formula to solve the following equations: • X2 + 3x – 4 = 0 • x(x – 4) = 8 • 2x2 – 4x – 3 = 0 • 9x2 + 12x + 4 = 0

  21. Application: 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.

  22. Example (cont.) Solution: From the formula, 125 = 0.006A² – 0.02A + 120. Rewriting, we get0.006A² – 0.02A – 5 = 0z where 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.

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