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next Key Stone Problem… Set 5 © 2007 Herbert I. Gross next Instructions for the Keystone Problem You will soon be assigned five problems to test whether you have internalized the material in Lesson 5 of our algebra course. The Keystone Illustration below is a prototype of the problems you'll be doing. Work out the problem on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve the problem. © 2007 Herbert I. Gross next As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. © 2007 Herbert I. Gross next Keystone Illustration for Lesson 5 A ball projected vertically upward at a speed of 160 feet per second, in the absence of air resistance, reaches a height of h feet at the end t seconds according to the rule: h = 400 – 16(t – 5)2 (a) How high up is the ball at the end of 7 seconds? © 2007 Herbert I. Gross Answer: 336 feet next Solution for Part a: h = 400 – 16 (4) (2) ( 7t 2– 5 )2 To solve part (a) we replace t by 7 in the formula. Using our PEMDAS agreement, we do what's inside the parentheses first, (7 – 5). We next replace (2)2 by 4 to obtain © 2007 Herbert I. Gross next Solution for Part a: h = 336 400 feet – 64 16 (4) And since we multiply 16 by 4 before we subtract, the equation becomes… Finally subtracting 64 from 400, the answer is 336. h is measured in feet so the answer to part (a) is 336 feet. © 2007 Herbert I. Gross next Keystone Illustration for Lesson 5 A ball projected vertically upward at a speed of 160 feet per second, in the absence of air resistance, reaches a height of h feet at the end t seconds according to the rule: h = 400 – 16(t – 5)2 (b) How high up is the ball at the end of 3 seconds? © 2007 Herbert I. Gross Answer: 336 feet next Solution for Part b: h = 400 – 16 ( -2) t3 2– 5 )2 The procedure for solving part (b) is exactly the same as the procedure for solving part (a). Namely, we replace t by 3 in the formula. Using our PEMDAS agreement, we do what's inside the parentheses first, (3 – 5). © 2007 Herbert I. Gross next Solution for Part b: h = + 2 336 400 feet – 16 64 ( 2) 4) (-2)2 means -2 × -2; which by our rule for multiplying two negative numbers is 4, so we next replace (-2)2 by 4 to obtain… We multiply 16 by 4, before we subtract. Subtracting 64 from 400, the answer is 336. h is measured in feet so the answer to part (b) is 336 feet. © 2007 Herbert I. Gross next Note • The fact that the product of two negative numbers is positive tells us that the square of any signed number is non-negative. More specifically, a signed number is either positive, negative or 0. © 2007 Herbert I. Gross next Note If we multiply a positive number by itself the product will be positive. If we multiply a negative number by itself the product will be positive. If we multiply 0 by itself the product will be 0. © 2007 Herbert I. Gross next Note • With respect to this example, if we replace t by 7, t – 5 = 2, and if we replace t by 3, t – 5 = -2. While +2 ≠ -2, (+2)2 = (-2)2. More generally if two numbers have the same magnitude their squares are equal. © 2007 Herbert I. Gross next Note • Based on the above note, it is incorrect to talk about the square root of a number. Every positive number has two square roots. In other words, for example, if (t – 5)2 = 4, t – 5 can be either 2 or -2. When we talk about the square root of a number we usually mean the positive square root of the number. However as we shall see in our next note, if we neglect the negative square root, we miss part of the answer to the present example. © 2007 Herbert I. Gross next Note • Because addition has nicer properties than subtraction, a good approach might be to rewrite h = 400 – 16 (t – 5)2 in the equivalent form h = 400 + -16 (t – 5)2 and then translate the formula into a verbal “recipe”. Start with t Step 1 Subtract 5 Step 2 Step 3 Square the result Step 4 Multiply by -16 Add 400 Step 5 Step 6 The answer is h. © 2007 Herbert I. Gross t t–5 (t – 5)2 -16(t – 5)2 400 + -16(t – 5)2 h = 400 + -16(t – 5)2 next Note • In the present example we found that when t = 3 or t = 7, h = 336. Let's now undo the above recipe and show why this occurred when t = 3 and t = 7. Recall that when we “undo” a recipe we start with the last step and replace each operation by the one that “undoes" it. In this case we see that… Start with 336 Step 1 Subtract 400 Step 2 Divide by -16 Step 3 Step 4 Take the (2) square root(s). Add 5 Step 5 The answer is t. Step 6 © 2007 Herbert I. Gross 336 336-64 – 400 -64+÷ 4 -16 +2 or √4 -2 2 +75 or or -23+ 5 t = 7 or 3 next • Notice that without the knowledge that a positive number has two square roots, step 4 would have read, “Take the square root of 4”, and the answer would have been only +2. Take thethe (2) square square root. root(s). Step 4 Take +2+√4 or 2 -2 and this would lead to our missing that t = 3 was also an answer. © 2007 Herbert I. Gross next The reason that two different values of t produce the same value for h is that the ball is at a given height twice, once on the way up, and once on the way down. © 2007 Herbert I. Gross 5s 400 feet 5s 4s 384 feet 6s 3s 336 feet 7s 2s 256 feet 8s 1s 144 feet 9s 0s 0 feet 10s next Note • The fact that the square of a signed number can never be negative gives us additional information that is contained in the formula. Namely… (t – 5)2 has to be non-negative; and the only time it can be 0 is if t – 5 = 0; that is, if t = 5. When t = 5, 16(t – 5)2 =0. © 2007 Herbert I. Gross next Note Since 16 (t – 5)2 can never be negative, whenever t ≠ 0, we are subtracting a positive number from 400. In other words if t represents any number other than 0, h, which equals 400 – 16(t – 5)2, is less that 400 feet. Therefore, the conclusion is that the ball reaches its greatest height (400 feet) when the time is 5 seconds. • © 2007 Herbert I. Gross next Note • Notice that 3 seconds is 2 seconds before the ball reaches its greatest height and that 7 seconds is 2 seconds after the ball reaches its maximum height. So if we think of 5 seconds as being our reference point, 3 seconds would be represented by -2, and 7 seconds would be represented by +2, if t = 5. In the above context -2 is just as meaningful as +2. © 2007 Herbert I. Gross next Summary The use of “profit and loss", “increase and decrease”, “below zero and above zero” give us good ways to visualize signed numbers. At the same time, however, they eliminate the need for us to deal with positive and negative numbers per se. That is we can talk about a $7 loss rather than a transaction of -$7, etc. © 2007 Herbert I. Gross next Moreover, even if we elect to use the terms “profit” and “loss” it would be difficult to give a physical reason as to why the product of two negative numbers is positive. For example we can interpret 3 × -2 = -6 by saying that, if we have a $2 loss three times, the net result is a $6 loss. However in looking at -3 × -2, it makes little sense to talk about a $2 loss “negative three” times or a $3 loss “negative two” times. © 2007 Herbert I. Gross next However in dealing with formulas as we did in the keystone exercise, we see why it is important to have a mathematical definition of signed numbers that transcends any particular real-life model. © 2007 Herbert I. Gross