2023 MWA Sampler

33 Algebra Sample Lesson Quadratic Equations Solving a quadratic equation by completeing the square Exponential and Quadratic Functions Unit 8 Lesson Plans 432 Objective: To solve a quadratic equation by completing the square. Materials: Completing the Square (Master 40), Solve Quadratic Equations Using Completing the Square (Master 41), plastic algebra tiles or paper algebra tiles made from Master 1 Vocabulary: completing the square Completing the Square We have learned about solving quadratic equations using square roots. Now we will learn how to complete the square so we can then solve the equation using square roots. Distribute a copy of Master 40 and algebra tiles to each student. Write on the board: x 2 – 6 x – 2 = 8 Build this equation using algebra tiles (shown below) . Now we will c omplete the square to solve the equation. The first step is to add 2 to both sides of the equation, to remove the constant from the left side. Use algebra tiles to model this. Record the steps on the board using symbols as well. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 – x – x – x – x – x – x – 1 – 1 x 2 – x – x – x – x – x – x x 2 = x 2 – 6 x – 2 = 8 +2 +2 x 2 – 6 x = 10 Now we will complete the square. This means we will use the tiles on the left side of the equation to form a square. Notice, we will likely have to add unit tiles to form the square. But, we must remember to add the tiles to both sides of the equation. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 – x – x – x – x – x – x – 1 – 1 x 2 – x – x – x – x – x – x x 2 = x 2 – 6 x – 2 = 8 +2 +2 x 2 – 6 x = 10 Notice, after isolating the constant, we are adding b 2 ( __ ) 2 = – 6 2 ( __ ) 2 = 9 unit squares to both sides of the equation. To complete the square, we always add b 2 ( __ ) 2 to both sides of the equation. Let’s now use symbols to write the left side of the equation as a “square” binomial. What are the side lengths of the square? ( x – 3) What is the area of the square? (( x – 3) 2 )) Now the equation is ( x – 3) 2 = 19. Since there is a square binomial, we are finished completing the square. Now we can use square roots to solve the equation. Continue to show the steps on the board. Remind students to write the plus or minus sign after taking the square root. Encourage students to find decimal approximations using calculators. The solutions are 3 +  19 < 7.36 and 3 –  19 < – 1.36. Refer to Master 40 to check the solutions (The solutions are the intersection points of the parabola y = x 2 – 6 x – 2 and the line y = 8.) Distribute Master 41 to all students. Work together as a class to model the first two examples using the completing the square process. Then ask students to work together to complete the third example. Monitor the students to assess student understanding. The solutions are – 2 +  6 < 0.45 and – 2 –  6 < – 4.45, 4 +  11 < 7.32 and 4 –  11 < 0.68, and 1 +  8 < 3.83 and 1 –  8 < – 1.83. As a class, work through the example at the top of the page. Ask students to complete the rest of the page. Skill Builders p. 364 ©Math TeachersPress, Inc. 432 Solving Quadratic Equations By Completing the Square 1. x 2 + 8 x = 2 ___________________________ 3. x 2 – 10 x + 3 = 0 ___________________________ 5. x 2 + 12 x + 5 = 0 ___________________________ 2. x 2 + 4 x – 1 = 0 ___________________________ 4. x 2 – 2 x – 3 = 0 ___________________________ 6. x 2 – 6 x – 2 = 0 ___________________________ Solveby completing the square.Give theexactvalues, thenapproximate the solutions to 2decimalplaces. Quadratic equations can be solved by “ completing the square ” and then by taking square roots. Solve: x 2 + 6 x – 10 = 0 Isolate the constant. x 2 + 6 x = 10 x 2 + 6 x + ____ = 10 + ____ Add ( ) 2 to both sides: ( ) 2 = ( ) 2 = 9 x 2 + 6 x + 9 = 19 Rewrite the quadratic expression as a binomial squared: ( x + 3) 2 = 19 Solve by using square roots. x + 3 = ±  19 x = – 3 ±  19 x = – 3 +  19 < 1.36 and x = – 3 –  19 < 7.36 x 2 1 1 1 1 1 1 1 1 1 x x x x x x 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 9 9 x + 3 x + 3 Exact values Approximate solutions x = – 4+  18 or x = – 4 –  18 x < 0.24 or x < – 8.24 x = – 2+  5 or x = – 2 –  5 x < 0.24 or x < – 4.24 x =5+  22 or x =5 –  22 x < 0.31 or x < 9.69 x = – 1 or x =3 x = – 6+  31 or x = – 6 –  31 x < – 0.43 or x < – 11.57 x =3+  11 or x =3 –  11 x < – 0.32 or x < 6.32