Solving Quadratic Equations
Word Problems -­‐ Answers Note: There are multiple ways to solve each of these problems. Now that we’re talking about real-­‐world applications, it’s a matter of choosing the best strategy you can for solving it. Unlike the first part of this test, any strategy you choose to use (factoring, completing the square, the quadratic equation, etc.) that leads you to a correct answer is acceptable. 1. A rectangle is 2 feet longer than it is wide. The area of the rectangle is 48 square feet. Write and solve an equation that can be used to find the width of the rectangle. What are the dimensions of the rectangle? Let’s say the width of the rectangle is x feet. That would mean the length of the rectangle is (x + 2) feet. Since the area is 48 square feet, that means the length times the width is 48. x(x + 2) = 48 x2 + 2x = 48 x2 + 2x – 48 = 0 (x + 8)(x – 6) = 0 x = –8 or x = 6 Since a width of –8 feet doesn’t make sense in the context of the problem, the width is 6 feet and the length is 8 feet. 2. A rectangle is twice as long as it is wide. The area of the rectangle is 32 square feet. Write and solve an equation that can be used to find the width of the rectangle. What are the dimensions of the rectangle? Let’s say the width of the rectangle is x feet. That would mean the length of the rectangle is 2x feet. Since the area is 32 square feet, that means the length times the width is 32. x(2x) = 32 2x2 = 32 x2 = 16 x = ±4 Since a width of –4 feet doesn’t make sense in the context of the problem, the width is 4 feet and the length is 8 feet. 3. A homeowner is planning an addition to her house. She wants the new family room to have an area of 192 square feet. The contractor says that the length needs to be 4 more feet than the width. What will the dimensions of the new room be? Let’s say the width of the room is x feet. That would mean the length of the room is (x + 4) feet. Since the area is 192 square feet, that means the length times the width is 192. x(x + 4) = 192 x2 + 4x = 192 x2 + 4x – 192 = 0 !4± 42 ! 4 ∙ 1 ∙ (!192)
x = x = x = x = –2 ± 14 x = 12 or x = –16 Since a width of –16 feet doesn’t make sense in the context of the problem, the width of the room must be 12 feet and the length 16 feet. 2 ∙ 1
4. !4± 784
2
!4±28
2
Tariq joins a recess kickball game. When he’s up, he kicks the ball. It takes 3.1 seconds for the ball to hit the ground. The trajectory of the ball can be represented by the quadratic equation h = –16t2 + bt + 0, where t represents time in seconds, and b is the initial upward speed of the ball (in feet per second) in the split second that Tariq kicks the ball. a) What is the initial upward speed of the ball? We know that the ball hits the ground after 3.1 seconds, and that its height at that moment is 0. So we can substitute those values into our equation: The initial upward speed of the ball is 49.6 feet per second. h = –16t2 + bt + 0 0 = –16(3.1)2 + b(3.1) + 0 3.1b – 153.76 = 0 b = 49.6 b) How high does the ball go into the air (its maximum height)? We know that the initial upward speed of the ball is 49.6 feet per second, so the full quadratic equation for the height of the ball is Since the ball will be at its highest point halfway through, and it hit the ground after 3.1 seconds, its highest point will be reached at 1.55 seconds. We can substitute that in for the time in seconds to find the ball’s height at that moment. The maximum height of the ball will be 38.44 feet above the ground. c) Why, in the quadratic equation, does c = 0 (in other words, why is it + 0 at the end of the left side of the equation, instead of + 3 or something)? What would it mean if it was + 3 (what would have been different about Tariq’s kick)? 5. h = –16t2 + 49.6t + 0 h = –16(1.55)2 + 49.6(1.55) + 0 h = –38.44 + 76.88 h = 38.44 feet In our quadratic equation, c = 0 because Tariq kicked the ball from the ground. If our quadratic equation had been h = –16t2 + bt + 3, it would mean he kicked the ball from 3 feet above the ground. (By the way, it also would mean that Tariq’s kick would not have had a starting speed of 49.6 feet per second, because a ball that is kicked from the ground and lands 3.1 seconds later and a ball that is kicked from 3 feet in the air and lands 3.1 seconds later – those would not be moving at the same speed.) Again on our playground, both Hannah and a kindergartner kick kickballs (from the ground) in the same instant. Hannah kicks hers with an initial upward speed of 46 feet per second. The kindergartner kicks his kickball with an initial upward speed of 20 feet per second. How many seconds after the kindergartner’s kickball hits the ground will Hannah’s kickball hit the ground? Round to the nearest hundredth of a second. Hannah’s kick: h1 = –16t2 + 46t Kindergartener’s kick: h1 = –16t2 + 20t When the ball is on the ground (both the moment the ball is kicked and the moment it hits the ground after flying through the air), its height is equal to zero. Let’s put 0 in for the heights (h1 and h2), and see what t would be. 6. Chloe’s kick: 2
0 = –16t + 46t –16t2 + 46t = 0 –2t(8t – 23) = 0 –2t = 0 or 8t – 23 = 0 23
t = 0 or t = 8 Kindergartener’s kick: 0 = –16t2 + 20t –16t2 + 20t = 0 –4t(4t – 5) = 0 –4t = 0 or 4t – 5 = 0 5
t = 0 or t = 4 So both Hannah’s and the kindergartner’s kicks begin at 0 seconds. Hannah’s ends at 2⅞ seconds, while the kindergartner’s kick ends at 1¼ seconds. Thus: 23
8
5
13
4
8
– = … Hannah’s kickball hits the ground 1⅝ seconds (1.625 seconds) after the kindergartner’s kickball. Rounded to the nearest hundredth of a second, that’s 1.63 seconds. Yet one more kickball problem: Kate kicks the ball (from the ground) towards the roof of the gym, with an initial upward speed of 45 feet per second. The height of the gym roof is 40 feet. Assuming she kicks it far enough, will Kate roof the ball? (In addition to solving the problem, explain how you can use the discriminant to solve this problem quickly). If Kate kicks the ball from the ground with a speed of 45 feet per second, the equation of the height of the ball by time is: h = –16t2 + 45t + 0 What we want to know is whether the ball will go over 40 feet in the air. In other words, is there any time at which the ball will go above 40 feet? Another way of saying the same thing is “if h is 40 feet, is there any value for t that will make the equation true”? 40 = –16t2 + 45t + 0 We can use the discriminant to find out the answer. If we use the discriminant and find that there will be two solutions, we know that the ball will go above 40 feet (there are two moments when it is exactly 40 feet above the ground – once when it is on its way up, and once on its way down. If we use the discriminant and find that there is only one solution, we know that the ball will reach exactly 40 feet – that would be its highest point (the maximum). And if we use the discriminant and find that there are no solutions, we know that there is no time at which the ball will reach 40 feet – no value for t will make the equation true. The quadratic equation we’ll use here is: 40 = –16t2 + 45t + 0 –40 –40 subtract 40 from both sides 0 = –16t2 + 45t – 40 So when we find a, b, and c to put into the discriminant, c is –40, not 0. In other words, the ball was kicked 40 feet below the spot we are now interested in. a = –16 b = 45 c = –40 b2 – 4ac 452 – 4 ·∙ (–16) ·∙ (–40) 2025 – 2560 –535 The discriminant is negative. If we put those numbers into the quadratic formula, we’d have to take the square root of a negative number. So there are no solutions – meaning that there is no value for t that makes that equation true. Kate’s kick cannot roof the ball.