Key Stone Problem… Key Stone Problem… next Set 5 © 2007 Herbert I. Gross

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. Instructions for the Keystone Problem next © 2007 Herbert I. Gross

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. next © 2007 Herbert I. Gross

next 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 Keystone Illustration for Lesson 5 next Answer: 336 feet © 2007 Herbert I. Gross (a) How high up is the ball at the end of 7 seconds? next

Solution for Part a: next © 2007 Herbert I. Gross h = 400 – 16 ( t – 5 ) 2 7 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). (2) 2 We next replace (2) 2 by 4 to obtain (4) next

Solution for Part a: next © 2007 Herbert I. Gross h = 400 – 16 (4) And since we multiply 16 by 4 before we subtract, the equation becomes… Finally subtracting 64 from 400, the answer is 336. next h is measured in feet so the answer to part (a) is 336 feet. feet

next 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 Keystone Illustration for Lesson 5 next Answer: 336 feet © 2007 Herbert I. Gross (b) How high up is the ball at the end of 3 seconds? next

Solution for Part b: next © 2007 Herbert I. Gross h = 400 – 16 ( t – 5 ) 2 3 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). ( - 2) 2 next

Solution for Part b: next © 2007 Herbert I. Gross h = 400 – 16 ( - 2) 2 ( - 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. next h is measured in feet so the answer to part (b) is 336 feet. ( + 4) Subtracting 64 from 400, the answer is feet next

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. next © 2007 Herbert I. Gross Note

next © 2007 Herbert I. Gross 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.

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. next © 2007 Herbert I. Gross 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. next © 2007 Herbert I. Gross 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 = (t – 5) 2 and then translate the formula into a verbal “recipe”. next © 2007 Herbert I. Gross Note Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Start with t Subtract 5 Square the result Multiply by - 16 Add 400 The answer is h.       t t – 5 (t – 5) (t – 5) (t – 5) 2 h = (t – 5) 2 next

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… next © 2007 Herbert I. Gross Note Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Start with 336 Subtract 400 Divide by - 16 Take the (2) square root(s). Add 5 The answer is t.       – ÷ - 16 √ or t = 7 or 3 next or or 3

next © 2007 Herbert I. Gross  and this would lead to our missing that t = 3 was also an answer. 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. next Step 4 Take the (2) square root(s).  √4 + 2 or Take the square root.

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. next © 2007 Herbert I. Gross 144 feet 256 feet 336 feet 384 feet 400 feet 0 feet 1s 2s 3s 4s 5s 6s 7s 8s 9s 10s0s 5s next

© 2007 Herbert I. Gross Note  (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. The fact that the square of a signed number can never be negative gives us additional information that is contained in the formula. Namely… next

© 2007 Herbert I. Gross 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. next

© 2007 Herbert I. Gross Note 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. 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. next

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 Summary

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. next © 2007 Herbert I. Gross 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. 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. next © 2007 Herbert I. Gross