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

Key Stone Problem…Key Stone Problem…

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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

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© 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.

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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

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Answer: 336 feet© 2007 Herbert I. Gross

(a) How high up is the ball at the end of 7 seconds?

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Solution for Part a:

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h = 400 – 16 ( t – 5 )27

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)

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Solution for Part a:

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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.

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64336

h is measured in feet so the answer to part (a) is 336 feet.

feet

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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

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Answer: 336 feet© 2007 Herbert I. Gross

(b) How high up is the ball at the end of 3 seconds?

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Solution for Part b:

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h = 400 – 16 ( t – 5 )23The 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

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Solution for Part b:

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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.

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h is measured in feet so the answer to part (b) is 336 feet.

(+4)

Subtracting 64 from 400, the answer is 336.

64336 feet

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• 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.

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Note

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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.

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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.

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Note

• Because addition has nicer properties than subtraction, a good approach might be

to rewrite h = 400 – 16 (t – 5)2 in theequivalent form h = 400 + -16 (t – 5)2 and then translate the formula into a verbal “recipe”.

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Note

Step 1Step 2Step 3Step 4Step 5Step 6

Start with tSubtract 5

Square the resultMultiply by -16

Add 400The answer is h.

tt – 5

(t – 5)2

-16(t – 5)2

400 + -16(t – 5)2

h = 400 + -16(t – 5)2

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• 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…

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Note

Step 1Step 2Step 3Step 4Step 5Step 6

Start with 336Subtract 400Divide by -16

Take the (2) square root(s).Add 5

The answer is t.

336336 – 400

-64 ÷ -16√4

2 + 5 or -2 + 5t = 7 or 3

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-64 +4

+2 or -2 7 or 3

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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.

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Step 4 Take the (2) square root(s). √4+2 or -2+2 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.

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144 feet

256 feet

336 feet

384 feet

400 feet

0 feet

1s

2s

3s

4s

5s

6s

7s

8s

9s

10s0s

5s

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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…

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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.

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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.

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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.

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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.

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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.

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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.

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