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Chapter 16 Quadratic Equations § 16.2 Solving Quadratic Equations by Completing the Square Completing the Square Solving a Quadratic Equation by Completing a Square 1) If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. 2) Isolate all variable terms on one side of the equation. 3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation). 4) Factor the resulting trinomial. 5) Use the square root property. Martin-Gay, Developmental Mathematics 3 Solving Equations Example Solve by completing the square. y2 + 6y = 8 y2 + 6y + 9 = 8 + 9 (y + 3)2 = 1 y+3=± 1=±1 y = 3 ± 1 y = 4 or 2 Martin-Gay, Developmental Mathematics 4 Solving Equations Example Solve by completing the square. y2 + y – 7 = 0 y2 + y = 7 y2 + y + ¼ = 7 + ¼ (y + ½)2 = 29 4 1 29 29 y 2 4 2 1 29 1 29 y 2 2 2 Martin-Gay, Developmental Mathematics 5 Solving Equations Example Solve by completing the square. 2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½ x2 + 7x + 49 4 =½+ 49 4 = 51 4 7 2 51 (x + ) = 2 4 x 7 51 51 2 4 2 7 51 7 51 x 2 2 2 Martin-Gay, Developmental Mathematics 6 § 16.3 Solving Quadratic Equations by the Quadratic Formula The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation. Martin-Gay, Developmental Mathematics 8 The Quadratic Formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions. b b 4ac x 2a 2 Martin-Gay, Developmental Mathematics 9 The Quadratic Formula Example Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1 9 (9) 4(11)( 1) 9 81 44 9 125 n 22 22 2(11) 2 95 5 22 Martin-Gay, Developmental Mathematics 10 The Quadratic Formula Example 1 2 5 Solve x + x – = 0 by the quadratic formula. 8 2 x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20 8 (8) 2 4(1)( 20) 8 64 80 8 144 x 2(1) 2 2 8 12 20 4 or , 10 or 2 2 2 2 Martin-Gay, Developmental Mathematics 11 The Quadratic Formula Example Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 6 (6) 4(1)(30) 6 36 120 6 84 x 2 2 2(1) 2 So there is no real solution. Martin-Gay, Developmental Mathematics 12 Graphs of Quadratic Equations Example y Graph y = 2x2 – 4. x y 2 4 1 –2 0 –4 –1 –2 –2 4 (–2, 4) (2, 4) x (–1, – 2) (1, –2) (0, –4) Martin-Gay, Developmental Mathematics 13 Intercepts of the Parabola Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y. Martin-Gay, Developmental Mathematics 14 Characteristics of the Parabola If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. b 2) the x-coordinate of the vertex is . 2a To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y. Martin-Gay, Developmental Mathematics 15 Graphs of Quadratic Equations Example Graph y = –2x2 + 4x + 5. Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex 4 is 1 y (0, 5) (1, 7) (2, 5) 2(2) x y 3 –1 2 5 1 7 0 5 –1 –1 (–1, –1) Martin-Gay, Developmental Mathematics (3, –1) x 16