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Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY x 2 8 x + 15 = 0 Consider the equation • Show (numerically) that x = 5 is a solution. . There is also another solution to the equation. Use the table to help find this second answer. x y x 2 8 x + 15 1 (1)2 – 8(1) + 15 8 2 (2)2 – 8(2) + 15 3 3 (3)2 – 8(3) + 15 0 4 (4)2 – 8(4) + 15 -1 5 (5)2 – 8(5) + 15 0 6 (6)2 – 8(6) + 15 3 Sample: (5)2 – 8(5) + 15 25 – 40 + 15 -15 + 15 0 Using the table above, plot the points (x, y) to sketch the graph of y = x 2 8 x + 15 . • Which two points represent the solutions to the original equation? Write down their coordinates. (3,0) and (5,0) • Where are both of these points located? Both points are on the x-axis. Here, we found both solutions to a quadratic equation using a method like trial-and-error. However, there are other (more efficient) methods for solving such quadratic equations: Part I: Graphs and Tables Part II: Factoring Part III: Quadratic Formula In all these cases, we use the following terms to describe the solutions to quadratic equations. x 2 6x + 8 = 0 ©2012, TESCCC 09/06/12 page 1 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 When we are dealing with general equations, then the answers are called When we are dealing with functions, y = x 2 6x + 8 such as , then the answers are called solutions or roots zeros or x-intercepts Here, we could say, x 2 6x + 8 = 0 “The solutions to are x = 2 and x = 4.” Here, we could say, y = x 2 6x + 8 “The zeros of the function occur at x = 2 and x = 4.” like this one, with x’s (but no y’s), ©2012, TESCCC 09/06/12 page 2 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY Part I: Graphs and Tables Use a graphing calculator to complete the chart. WINDOW : Factored Equation Ex. Expanded Equation y = (x – 2)(x + 2) 2 y = ( x 2)( x + 2) y = x + 2x – 2x – 4 Graph (Sketch) Xmin= -9.4 Xmax= 9.4 Zeros of Function x = -2 x=2 Ymin= -12 Ymax= 12 Check (with Table) x -2 0 2 y 0 -4 0 x -1 4 y 0 0 x -3 -6 y 0 0 x 5 y 0 x -2 2.5 y 0 0 x -4 1.5 y 0 0 y = x2 – 4 1) 2) y = ( x 4)( x + 1) y = x2 – 3x – 4 y = ( x + 3)( x + 6) y = x2 + 9x + 18 x = -1 x=4 x = -3 x = -6 y = ( x 5)( x 5) 3) , or y = ( x 5) 2 y = x2 – 10x + 25 4) y = 2( x 2.5)( x +y2=) -2x2 + x + 10 5) y = 2(x – 1.5) (x + 4), or y = (2x – 3)(x + 4) ©2012, TESCCC y = 2 x 2 + 5 x 12 09/06/12 x=5 x = -2 x = 2.5 x = -4 x = 1.5 page 3 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 6) ©2012, TESCCC None (no real solutions) y = x2 + 4 09/06/12 x -1 0 1 y 5 4 5 page 4 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY Questions: 7) How is each factored equation related to the zeros of its related function? Answers may vary. Sample: The factors have numbers that are the opposites of the zeros. Sample: If n is a zero of the function, then (x – n) is a factor. y = ( x 5)( x 5) 8) Equation #3 used the same factor twice: . How does this affect the graph of its related function? Answers may vary. Sample: The graph only has one x-intercept. Sample: The graph touches (but does not cross) the x-axis at the point. y = 2( x 2.5)( x + 2) 9) Equation #4 used a negative coefficient in front of its factors: this affect the graph of its related function? Answers may vary. Sample: The negative number made the graph point down instead of up. Sample: The negative coefficient reflected the graph over the x-axis. . How does Think About It: The Fundamental Theorem of Algebra states that a polynomial equation of degree n has precisely n complex roots. • What is the degree of a quadratic equation? 2 • How many complex roots should it have? It should have two complex roots. Study the graphs of the quadratic functions below. Explain how their related equations can each have “two complex roots.” Explain: Two real roots. The two solutions are where it crosses the x-axis and can be rational or irrational. ©2012, TESCCC Explain: One real “double root.” The two solutions are equal to the same number. This is why it is called a double root. 09/06/12 Explain: No real solutions. Solutions cannot be real numbers because it does not cross the axis. These solutions involve imaginary numbers, but are still considered “complex.” page 5 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 ©2012, TESCCC 09/06/12 page 6 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY Sample Problems Complete the tables to determine the zeros of each quadratic function. A) B) y = x 2 2x 3 x -2 -1 0 1 2 3 4 y 5 0 -3 -4 -3 0 5 Zeros: x = -1, x = 3 f ( x ) = 2x 2 9 x + 7 x 1 1.5 … 3 3.5 4 4.5 y 0 -2 … -2 0 3 7 Zeros: x = 1, x = 3.5 Use a calculator to sketch the graph of each function. Then tell where the x-intercepts occur. y = x 2 4x 5 y = 2 x 2 + 24 x + 72 y = x 2 3x + 5 C) Graph: D) Graph: E) Graph: x-intercepts: x = -1, x = 5 x-intercepts: x = -6 (double root) x-intercepts: None (no real solutions) F) Write down the zeros of the quadratic function graphed here. Work backward and use the zeros to write the two factors of the function. Multiply the factors to find an “expanded” equation for the function. ©2012, TESCCC Zeros: x = -3 and x = 2 Factors: (x + 3) and (x – 2) Function: y = x2 + x – 6 09/06/12 page 7 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 ©2012, TESCCC 09/06/12 page 8 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY Part II: Factoring Follow these steps to solve quadratic equations by factoring. 1. Set each equation equal to zero. 2. Factor. 3. Set each factor equal to zero and solve. 4. Check solutions by graphing the representative function and comparing zeros of the function. NOTE: Factoring cannot be used to find irrational or imaginary solutions. Sample Problems A) x2 x = 6 x2 – x – 6 = 0 (x – 3)(x + 2) = 0 {-2, 3} B) 2x 2 = 5 9 x 2x2 + 9x – 5 = 0 (x + 5)(2x – 1) = 0 {-5, ½ } C) 2 x 2 4 x = 8 x + 16 2x2 + 4x – 16 = 0 2(x + 4)(x – 2) = 0 {-4, 2} ©2012, TESCCC 09/06/12 page 9 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 Solving Quadratic Equations by Graphs and Factoring KEY PRACTICE PROBLEMS Find the zeros of each function using tables and graphs. y= 1) Table: 1 2 x 1 2 3 4 5 6 y = x 2 14 x + 49 2) Graph (sketch): 3) Graph (sketch): Zeros: x = 2.75, x = 6 Zeros: x = 7 (double root) y 7.5 4 1.5 0 -0.5 0 Zeros: x = 4, x = 6 4) y = 4 x 2 35 x + 66 x 2 5 x + 12 Write down the zeros of the quadratic function graphed here. Work backward and use the zeros to write the two factors of the function. Multiply the factors to find an “expanded” equation for the function. Zeros: x = -4 and x = 1 Factors: (x + 4) and (x – 1) Function: y = x2 + 3x – 4 Find the solutions to the quadratic equations by factoring. Check all possible solutions. 5) x 2 + 6x + 9 = 0 6) {-3} double root 7) 2x 2 + 7x + 3 = 0 − {-3, 1 2 ©2012, TESCCC x 2 + 3 x = 10 {-5, 2} 8) } 3x 2 + x = 2 {-1, 09/06/12 2 3 } page 10 of 11 Algebra 2 HS Mathematics Unit: 05 Lesson: 02 ©2012, TESCCC 09/06/12 page 11 of 11