Chapter 16 Quadratic Equations Chapter Sections 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving Quadratic Equations by Completing the Square 16.3 – Solving Quadratic Equations by the Quadratic Formula 16.4 – Graphing Quadratic Equations in Two Variables 16.5 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions Martin-Gay, Developmental Mathematics 2 § 16.1 Solving Quadratic Equations by the Square Root Property Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then a b Martin-Gay, Developmental Mathematics 4 Square Root Property Example Solve x2 = 49 x 49 7 Solve 2x2 = 4 x2 = 2 x 2 Solve (y – 3)2 = 4 y 3 4 2 y=32 y = 1 or 5 Martin-Gay, Developmental Mathematics 5 Square Root Property Example Solve x2 + 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number. Martin-Gay, Developmental Mathematics 6 Square Root Property Example Solve (x + 2)2 = 25 x 2 25 5 x = 2 ± 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7 Martin-Gay, Developmental Mathematics 7 Square Root Property Example Solve (3x – 17)2 = 28 3x – 17 = 28 2 7 3x 17 2 7 17 2 7 x 3 Martin-Gay, Developmental Mathematics 8 § 16.2 Solving Quadratic Equations by Completing the Square Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples). Martin-Gay, Developmental Mathematics 10 Completing the Square Example What constant term should be added to the following expressions to create a perfect square trinomial? x2 – 10x add 52 = 25 x2 + 16x add 82 = 64 x2 – 7x 2 49 7 add 4 2 Martin-Gay, Developmental Mathematics 11 Completing the Square Example We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Martin-Gay, Developmental Mathematics 12 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 13 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 14 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 15 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 16 § 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 18 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 19 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 20 The Quadratic Formula Example Solve 1 8 x2 +x– 5 2 = 0 by the quadratic formula. 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 21 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 22 The Discriminant The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively. Martin-Gay, Developmental Mathematics 23 The Discriminant Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions. Martin-Gay, Developmental Mathematics 24 Solving Quadratic Equations Steps in Solving Quadratic Equations 1) If the equation is in the form (ax+b)2 = c, use the square root property to solve. 2) If not solved in step 1, write the equation in standard form. 3) Try to solve by factoring. 4) If you haven’t solved it yet, use the quadratic formula. Martin-Gay, Developmental Mathematics 25 Solving Equations Example Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = -3, c = 1 3 (3) 4(1)(1) 3 9 4 3 5 x 2 2 2(1) 2 Martin-Gay, Developmental Mathematics 26 Solving Equations Example Solve the following quadratic equation. 5 2 1 m m 0 8 2 5m 2 8m 4 0 (5m 2)( m 2) 0 5m 2 0 or m 2 0 2 m or m 2 5 Martin-Gay, Developmental Mathematics 27 § 16.4 Graphing Quadratic Equations in Two Variables Graphs of Quadratic Equations We spent a lot of time graphing linear equations in chapter 3. The graph of a quadratic equation is a parabola. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola. Martin-Gay, Developmental Mathematics 29 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 30 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 31 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 32 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 33 § 16.5 Interval Notation, Finding Domain and Ranges from Graphs, and Graphing Piecewise-Defined Functions Domain and Range Recall that a set of ordered pairs is also called a relation. The domain is the set of x-coordinates of the ordered pairs. The range is the set of y-coordinates of the ordered pairs. Martin-Gay, Developmental Mathematics 35 Domain and Range Example Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)} • Domain is the set of all x-values, {4, –4, 2, 10} • Range is the set of all y-values, {9, 3, –5} Martin-Gay, Developmental Mathematics 36 Domain and Range y Example Domain Find the domain and range of the function graphed to the right. Use interval notation. Domain is [–3, 4] x Range Range is [–4, 2] Martin-Gay, Developmental Mathematics 37 Domain and Range y Example Find the domain and range of the function graphed to the right. Use interval notation. Range x Domain is (– , ) Range is [– 2, ) Domain Martin-Gay, Developmental Mathematics 38 Domain and Range Example Find the domain and range of the following relation. Input (Animal) Output (Life Span) • Polar Bear 20 • Cow • Chimpanzee 15 • Giraffe • Gorilla 10 • Kangaroo • Red Fox 7 Martin-Gay, Developmental Mathematics 39 Domain and Range Example continued Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox} Range is {20, 15, 10, 7} Martin-Gay, Developmental Mathematics 40 Graphing Piecewise-Defined Functions Example 3x 2 if x 0 . Graph f (x) x 3 if x 0 Graph each “piece” separately. Values 0. x f (x) = 3x – 1 x f (x) = x + 3 0 – 1(closed circle) 1 4 2 5 3 6 –1 – 4 Values > 0. –2 – 7 Continued. Martin-Gay, Developmental Mathematics 41 Graphing Piecewise-Defined Functions Example continued y x f (x) = 3x – 1 0 – 1(closed circle) –1 – 4 –2 – 7 (3, 6) Open circle (0, 3) (0, –1) x f (x) = x + 3 1 4 2 5 3 6 x (–1, 4) (–2, 7) Martin-Gay, Developmental Mathematics 42

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