Solutions for Quadratic Equations and Applications

Transcription

1 Solutions for Quadratic Equations and Applications I. Souldatos January 14, Solutions Disclaimer: Although I tried to eliminate typos multiple times, there might still be some typos around. If you find any of them, please let me know. Problem Write the function (5x + 3)(x 5) in standard form ax 2 + bx + c. Multiply out: (5x + 3)(x 5) = 5x 2 25x + 3x 15 = 5x 2 22x Solve the equation (5x + 3)(x 5) = 0. First way: Using the factored form set both parentheses equal to 0. 1

2 (5x + 3)(x 5) = 0 (5x + 3) = 0 or (x 5) = 0 5x = 3 or x = 5 x = 3 5 or x = 5 Second way: Use the standard form 5x 2 22x 15 = 0 and the quadratic formula. x = b ± b 2 4ac = 22 ± ( 15) 2 5 = 22 ± ± 28 = x = or x = x = 1.6 or x = 5 Notice that the solutions are the same no matter what way you use. Grade yourself: One point for part (1), and one point for each of the solutions. Subtotal: / Write the function (x 87) 2 60 in standard form ax 2 + bx + c. Page 2 of 9

3 Multiply out: (x 87) 2 60 = x 2 2 x = x 2 174x Solve the equation (x 87) 2 60 = 0. First way: Use the vertex form and follow the steps: (a) Move 60 to the right, (b) take square root of both sides, and (c) move 87 to the right. (x 87) 2 60 = 0 (x 87) 2 = 60 x 87 = ± 60 x = 87 ± 60 Second way: Use the standard form x 2 174x = 0 and the quadratic formula. x = b ± b 2 4ac = 174 ± = 174 ± In this case, the solutions look different, but verify that they are the same. Grade yourself: One point for part (1), and two points for part (2). Subtotal: /3 Page 3 of 9

4 Complete the square by writing 3x x + 1 in the form a(x h) 2 + k. h = b = = 36 6 = 6 Plug h = 6 into the function 3x x + 1 to find k. k = 3( 6) ( 6) + 1 = ( 6) + 1 = 107 So, 3x x + 1 = 3(x + 6) Observe that in the parenthesis the sign of h = 6 changed to +6, while outside the parenthesis the sign of k = 107 stayed the same. 2. Solve the equation 3x x + 1 = 0. First way: Use the vertex form and follow the steps: (a) Move 107 to the right, (b) divide by 3, (c) take square root of both sides, and (d) move 6 to the right. Page 4 of 9

5 3(x + 6) = 0 3(x + 6) 2 = 107 (x + 6) 2 = x + 6 = ± x = 6 ± 3 Second way: Use the standard form 3x x + 1 = 0 and the quadratic formula. x = b ± b 2 4ac = 36 ± = 36 ± In this case too, the solutions look different, but verify that they are the same. Grade yourself: One point for finding h, one point for finding k, and two points for part (2). Subtotal: / Find the discriminant of the equation 4x 2 + 2x + 7 = 0. D = b 2 4ac = = = 108 < 0 Page 5 of 9

6 2. Use the discriminant to say whether the equation has two (distinct) solutions, one solution, or no solutions. Since D < 0, there are no solutions. Grade yourself: One point for the discriminant, and one point for deciding how many solutions there are. Subtotal: / Factor the expression 2x 2 + 6x. The easiest way here is to factor out x. 2. Solve the equation 2x 2 + 6x = 0. 2x 2 + 6x = x(2x + 6) First way: Using the factored form set both factors equal to 0. x(2x + 6) = 0 x = 0 or (2x + 6) = 0 x = 0 or 2x = 6 x = 0 or x = 3 Second way: Use the standard form 2x 2 + 6x = 0 and the quadratic formula. Observe that there is no c. In this case c = 0. Page 6 of 9

7 x = b ± b 2 4ac = 6 ± = 6 ± 6 4 x = 6 6 or x = x = 3 or x = 0 Notice again that the solutions are the same no matter what way you use. Grade yourself: One point for factoring, and one point for each of the solutions. Subtotal: / Solve the quadratic equation (3x + 5)(x 2) = 7. Hint: If the right side of the equation was 0, then the equation would be in factored form. First step is to move the 7 to the left side. Can you find the next steps? (3x + 5)(x 2) = 7 (3x + 5)(x 2) 7 = 0 3x 2 6x + 5x 10 7 = 0 3x 2 x 17 = 0 x = 1 ± ( 17) 2 3 x = 1 ± Page 7 of 9

8 Grade yourself: One point for moving 7 to the left and multiplying out, one point for combining like terms, and one point for the quadratic formula. Subtotal: /3 Problem 2. The height of an object at time t (in seconds) is given by the equation (a) What is the initial height of the object? h(t) = 2(t 3) This problem is mainly conceptual. You have to understand what the different pieces are asking. Initial height means the height when t = 0. So, plug t = 0 into the height equation. h(0) = 2 ( 3) = = 6 Grade yourself: One point for setting t = 0, and one point for finding h(0). Subtotal: /2 (b) When does the object hit the ground? When the object hits the ground, the height is 0. So, set h(t) = 0 and solve. 2(t 3) = 0 2(t 3) 2 = 24 (t 3) 2 = 12 t 3 = ± 12 t = 3 ± 12 The decimal approximations for 3 ± 12 are 6.46 and Since t > 0, we keep only the first solution. Grade yourself: One point for setting h(t) = 0, and two points for solving for t. Subtotal: /3 Page 8 of 9

9 (c) What is the maximum height? We need the vertex of h(t). Since h(t) is already in the vertex form, we just read-off h = 3 and k = 24. The maximum height is equal to the y-component of the vertex. So, maximum height is 24. Grade yourself: One point for finding the vertex, and one point for the maximum height. Subtotal: /2 (d) When does the object reach maximum height? This is asking the t-component of the vertex. So, when t = 3 seconds, the object attains the maximum height. Grade yourself: One point for this part. Subtotal: /1 Total for problem 2: /8 Page 9 of 9