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Chapter 9

Chapter 9. Section 2. Solving Quadratic Equations by Completing the Square. Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1. Solve quadratic equations by completing the square when the coefficient of the second-degree term is not 1.

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Chapter 9

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  1. Chapter 9 Section 2

  2. Solving Quadratic Equations by Completing the Square Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1. Solve quadratic equations by completing the square when the coefficient of the second-degree term is not 1. Simplify the terms of an equation before solving. Solve applied problems that require quadratic equations. 9.2 2 3 4

  3. Objective 1 Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1. Slide 9.2-3

  4. The methods studied so far are not enough to solve the equation x2 + 6x + 7 = 0. Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1. If we could write the equation in the form (x + 3)2 equals a constant, we could solve it with the square root property discussed in Section 9.1. To do that, we need to have a perfect square trinomial on one side of the equation. Recall from Section 6.4 that a perfect square trinomial has the form x2 + 2kx + k2 or x2– 2kx + k2, where krepresents a number. Slide 9.2-4

  5. Complete each trinomial so that it is a perfect square. Then factor the trinomial. x2 + 12x + ______ x2– 14x + ______ EXAMPLE 1 Creating Perfect Square Trinomials Solution: Factored as: Factored as: Slide 9.2-5

  6. Solve quadratic equations by completing the square when the coefficient of the second-degree term is 1. (cont’d) The process of changing the form of the equation from x2 + 6x + 7 = 0 to (x + 3)2 = 2 is called completing the square. Completing the square changes only the form of the equation. Completing the square not only provides a method for solving quadratic equations, but also is used in other ways in algebra (finding the coordinates of the center of a circle, finding the vertex of a parabola, and so on). Slide 9.2-6

  7. Solve x2– 20x + 34 = 0. EXAMPLE 2 Rewriting an Equation to Use the Square Root Property Solution: Slide 9.2-7

  8. Solve x2 + 4x = 1. EXAMPLE 3 Completing the Square to Solve a Quadratic Equation Solution: Slide 9.2-8

  9. Objective 2 Solve quadratic equations by completing the square when the coefficient of the second-degree term is not 1. Slide 9.2-9

  10. Solving a Quadratic Equation by Completing the Square Step 1:Be sure the second-degree term has a coefficient of 1.If the coefficient of the second-degree term is 1, go toStep 2.If it is not 1, but some other nonzero number a, divide each side of the equation by a. If the solutions to the completing the square method are rational numbers, the equations can also be solved by factoring. However, the method of completing the square is a more powerful method than factoring because it allows us to solve any quadratic equation. Solve quadratic equations by completing the square when the coefficient of the second-degree term is not 1. Step 2:Write in correct form.Make sure that all variable terms are on the one side of the equation and that all constant terms are on the other. Step 3:Complete the square.Take half of the coefficient of the first degree term, and square it. Add the square to both sides of the equation. Factor the variable side and combine like terms on the other side. Step 4:Solvethe equation by using the square root property. Slide 9.2-10

  11. EXAMPLE 4 Solving a Quadratic Equation by Completing the Square Solution: Solve 4x2+ 8x -21 = 0. Slide 9.2-11

  12. Solve 3x2 + 5x− 2 = 0. EXAMPLE 5 Solving a Quadratic Equation by Completing the Square Solution: Slide 9.2-12

  13. Solve 5v2 + 3v + 1 = 0. EXAMPLE 6 Solving a Quadratic Equation by Completing the Square Solution: No solution, because the square root of a negative number is not a real number, therefore the solution set is Ø. Slide 9.2-13

  14. Objective 3 Simplify the terms of an equation before solving. Slide 9.2-14

  15. Solve (x + 6)(x + 2) = 1. EXAMPLE 7 Simplifying the Terms of an Equation before Solving Solution: Slide 9.2-15

  16. Objective 4 Solve applied problems that require quadratic equations. Slide 9.2-16

  17. Suppose a ball is projected upward with an initial velocity of 128 ft per sec. Its height at time t (in seconds) is given by s = –16t 2 + 128t, where s is in feet. At what times will the ball be 48 ft above the ground? Give answers to the nearest tenth. EXAMPLE 8 Solving a Velocity Problem Solution: Let s = 48 ft. The ball will be at a height of 48 ft from the ground at the times of 0.4 seconds and 7.6 seconds. Slide 9.2-17

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