R. l Basic Algebra: Real Numbers and Inequalities

PREREQUISITES

1. There are no prerequisites for this section other than some high school

algebra and geometry; however, if the material presented in this section

is new to you, it would be a good idea to enroll in a precalculus course.

This section is intended to be a review.

PREREQUISITE QUIZ

1. Orientation quizzes A and B in the text will help you evaluate your

preparation for this section and this course.

GOALS

1. Be able to factor and expand common mathematical expressions .

2. Be able to complete a square.

3. Be able to use the quadratic formula .

4. Be able to solve equations and inequalities .

STUDY HINTS

1. Common identities. Know how to factor a^{2} - b^{2} . It is a
good idea to

memorize the expansion of (a + b)^{2} and (a + b)^{3} . Note
that (a - b)^{2}

can be obtained by substituting -b for can
be similarly

expanded. These identities are useful for computing limits in Section 1.2

and Chapter 11 .

2. Factoring. This is a technique that is learned best through practice. A

good starting point is to find all integer factors of the last term (the

constant term). Once you find a factor for the original polynomial, use

long division to find a simpler polynomial to factor. This will be important

for partial fractions in Chapter 10 and for computing limits.

3. Completing the square. Don't memorize the formula. Practice until you

learn the technique. Note that adding ( b/ 2a )^{2} to x^{2} +
bx/a forms a

perfect square . This technique will be very important for integration

techniques introduced in Chapter 10.

4. Quadratic formula. - It is recommended that you memorize this formula. It.

is used in many applications in various disciplines such as engineering,

economics, medicine, etc. This formula may also be used to solve equations

of the form Ax^{4} + Bx^{2} + C = 0 by solving for y = x^{2}
and taking

square roots to get x .

5. Square roots. Note that, unless otherwise stated, square roots are understood

to be nonnegative.
is equal to zero.

6. Inequalities. It is essential to have a good handle on manipulating

inequalities. Without this, you will not have a good understanding of

some of the basic theory of calculus. Don't forget to reverse the direction

of the inequality sign when you multiply by a negative number.

SOLUTIONS TO EVERY OTHER ODD EXERCISE

1.
is a rational number. Since the denominator cannot

be reduced to one , it is neither a natural number nor an integer.

5. ( a - 3 ) ( b + c ) - ( a c + 2 b ) = ( a b - 3 b + a c - 3 c ) - ( a c + 2 b
) = a b - 5 b - 3 c .

9. We can use Example 2 with b replaced by -b t o get

. Alternatively , write

13. We know t h a t (x + a) (x + b) = x^{2} + (a + b)x + ab . The factor
s of 6

are
, and
. By choosing a = 2 and b = 3 , we

get a + b = 5 . Thus, x^{2} + 5 x + 6 = ( x + 2 ) ( x + 3 ) .

17. First we factor out 3 to get 3(x^{2} - 2x - 8) . We know that the
fact-

ors of -8 are
, and
. As in Exercise 13, we

look for a and b so that (a + b)x is the middle term. In this case,

a = -4 and b = 2 . Thus, 3x^{2} - 6x - 24 = 3(x - 4 ) ( x + 2 ) .

21. 2(3x - 7) - (4x - 10) = 0 simplifies to (6x - 14) - (4x - 10) =

2x - 4 = 0 , i . e . , 2x = 4 . Dividing by 2 yields x = 2 .

25. The right-hand side is

, which is the left-hand side.

29. (a) By factoring, we get x^{2} + 5x + 4 = (x + 4) (x + 1) = 0 . If
any

factor equals zero, the equation is solved. Thus, x = -4 or x = -1 .

(b) By using the method of completing the square, we get 0 = x^{2} + 5x
+ 4 =

. Rearrangement

yields
, and taking square roots gives

. Again, x = -4 or -1 .

(c) a = 1 , b = 5 , and c = 4 , so the quadratic formula

33. We use the quadratic formula with a = -1 , b = 5 , and
c = 0.3 .

This gives
. These are the

two solutions for x .

37. x^{2} + 4 = 3x^{2} - x is equivalent to 0 = 2x^{2} -
x - 4 . Using the quadratic

formula with a = 2 , b = -1 , and c = -4 , we get

41. We apply the quadratic formula with a = 2 ,
, and to

get . Thus, the only solution is

45. We add b to both sides to obtain a + c > 2c . Then we subtract c to

get a > c .

49. b(b + 2) > (b + 1) (b + 2) is equivalent to
.

Subtracting b^{2} + 2b from both sides leaves 0 > b + 2 . Subtracting 2

yields - 2 > b .

53. (a) Dividing through by a in the general quadratic equation yields

x^{2} + (b/a)x + c/a = 0 . Add and subtract
to get

Taking

square roots gives , and finally

(b) From the quadratic formula, we see that there are no solutions if

b^{2} - 4ac < 0 . If b^{2} - 4ac > 0 , there are two distinct
roots.

However, if b^{2} - 4ac = 0 , there are two roots, which both equal

-b/2a . This only occurs if b^{2} = 4ac .

SECTION QUIZ

1. Factor:

2. Apply the expansion of (a + b)^{3} t o expand (3x - 2)^{3} .

3. Use the quadratic formula t o solve x^{5} + 3x^{3} - 5x = 0 .

4. Sketch the solution of

(a) x^{2} + 3x + 2 < 0

(b) x^{2} + 3x + 2 ≥ 0

5. Find the solution set of .

6. The firs t King of the Royal Land of Mathernatica has decreed t h a t the

firs t young lady t o answer the following puzzle shall rule at his side .

The puzzle is t o compute the product of a l l solutions to

x^{3} + 2x^{2} - x - 2 = 0 . Then, divide by the length of the
finite interval

f o r which x^{3} + 2x^{2} - x - 2 ≥ 0 . What answer would make
a lady Queen

of Mathernatica ?*

ANSWERS TO SECTION QUIZ

*Dear Reader: I realize that many of you h a t e math but a r e forced t o comp

lete this course for graduation, Thus, I have attempted t o maintain interest

with " entertaining " word problems. They are not meant t o be insulting

to your intelligence . Obviously, most of the situations will never

happen; however, calculus has several practical uses and such examples are

found throughout Marsden and Weinstein's text . I would appreciate your

comments on whether my "unusual" word problems should be kept for the next

edition .