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Ups and Downs: Solving Quadratic Equations with the Quadratic Formula

In this activity, we explore solving quadratic equations using the quadratic formula. We look at a scenario involving a soccer ball's flight path modeled by a quadratic function to estimate its distance and the number of times it reaches a certain height. By discussing vocabulary and working through examples, we learn how to find solutions to quadratic equations, identify zeros of a function, and understand the implications of the number of solutions graphically. The activity concludes with sample problems and homework for further practice.

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Ups and Downs: Solving Quadratic Equations with the Quadratic Formula

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  1. 5-Minute Check on Activity 4-5 Factor the following quadratic polynomials: y = x2 – 6x + 8 y = x2 – 2x – 24 y = 4x2 – 8x -20 = x2 – 2x – 35 24 = x2 – 5x = (x – 2)(x – 4) = (x – 6)(x + 4) = 4x(x – 2) = x2 – 2x – 15 = (x – 5)(x + 3) = x2 – 5x – 24 = (x – 8)(x + 3) Click the mouse button or press the Space Bar to display the answers.

  2. Activity 4 - 6 Ups and Downs

  3. Objectives • Use the quadratic formula to solve quadratic equations • Identify solutions of a quadratic equation with points on the corresponding graph • Determine the zeros of a function

  4. Vocabulary • Quadratic formula – an equation that provides solutions to quadratic equations in standard form.

  5. Activity Suppose a soccer goalie punted the ball in such a way as to kick the ball as far as possible down the field. The height of the ball above the field as a function of time can be approximated by Y = -0.017x2 + 0.98x + 0.33 Where y represents the height of the ball (in yards) and x represents the horizontal distance in yards down the field from where the goalie kicked the ball. In this situation, the graph of the function is the actual path of the flight of the soccer ball. The graph of this function appears below:

  6. y Activity cont Y = -0.017x2 + 0.98x + 0.33 The graph of this function appears below: Use the graph to estimate how far downfield from the point of contact the soccer ball in 10 yards above the ground. How often during its flight does this occur? Write a quadratic equation to determine when the ball is 10 yards above the ground. x 20 15 10 Appears to happen twice; about 15 and 45 5 10 20 30 40 50 60 70 -0.017x2 + 0.98x + 0.33 = 10

  7. y Activity cont Y = -0.017x2 + 0.98x + 0.33 The graph of this function appears below: Put into Standard form for a quadratic: What does the graph indicate for the number of solutions? How can we solve this graphically on our calculator? x 20 -0.017x2 + 0.98x – 9.67 = 0 15 10 5 Two as said before 10 20 30 40 50 60 70 • Draw y = 10; Intersection • Draw only standard form; Zeros

  8. Quadratic Formula For a quadratic equation in standard form: ax2 + bx + c = 0, a ≠ 0, then its solutions are given by The  in the formula indicates that there are two solutions, one in which –b is added to the square root and the other in which the square root is subtracted from –b. -b  b2 – 4ac x = ---------------------- 2a

  9. Quadratic Formula Example 1 Solve x2 – x – 6 = 0 1) Identify parts: a = 1; b = -1 and c = -6 -b  b2 – 4ac x = ---------------------- 2a -(-1)  (-1)2 – 4(1)(-6) x = ------------------------------- 2(1) 1  1 + 24 1  5 x = ------------------ = --------- 2 2 2) Plug in: 3) Simplify: 4) Solve: x = 3 or x = -2

  10. Quadratic Formula Example 2 Solve 6x2 – x = 2 0) Standard Form: 6x2 – x – 2 = 0 1) Identify parts: a = 6; b = -1 and c = -2 -b  b2 – 4ac x = ---------------------- 2a -(-1)  (-1)2 – 4(6)(-2) x = ------------------------------- 2(6) 1  1 + 48 1  7 x = ------------------ = --------- 12 12 2) Plug in: 3) Simplify: 4) Solve: x = 2/3 or x = -1/2

  11. Quadratic Formula Example 3 Solve x2 + x + 6 = 0 1) Identify parts: a = 1; b = 1 and c = 6 -b  b2 – 4ac x = ---------------------- 2a -(1)  (1)2 – 4(1)(6) x = ------------------------------- 2(1) 1  1 - 24 1  -23 x = ------------------ = ------------ 2 2 2) Plug in: 3) Simplify: 4) Solve: no real solutions

  12. Activity Revisited Solve -0.017x2 + 0.98x + 0.33 = 10 0) Standard Form: -0.017x2 + 0.98x – 9.67 = 0 1) Identify parts: a = -0.017; b = 0.98 and c = -9.67 -b  b2 – 4ac x = ---------------------- 2a -(0.98)  (0.98)2 – 4(-0.017)(-9.67) x = ------------------------------------------------ 2(-0.017) -0.98  0.9604 – 0.6576 x = ------------------------------------- -0.034 2) Plug in: 3) Simplify: 4) Solve: x ≈ 12.634 or x ≈ 45.013

  13. Summary and Homework • Summary • The quadratic equation will provide solutions to all quadratic equations in standard form • If the value under the square root in negative, then the solutions are not real (complex #) • Homework • pg 449 – 453; problems 1-4, 6 -b  b2 – 4ac x = ---------------------- 2a

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