REVIEW OF ALGEBRA - Stewart Calculus

REVIEW OF ALGEBRA Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. ARITHMETIC OPERATIONS

The real numbers have the following properties: abba ab ba a b c a b c ab c ab ac

abc abc

(Commutative Law) (Associative Law) (Distributive law)

In particular, putting a 1 in the Distributive Law, we get b c 1b c 1b 1c and so b c b c EXAMPLE 1

(a) 3xy4x 34x 2y 12x 2y (b) 2t7x 2tx 11 14tx 4t 2x 22t (c) 4 3x 2 4 3x 6 10 3x If we use the Distributive Law three times, we get a bc d a bc a bd ac bc ad bd This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a bc d In the case where c a and d b, we have a b2 a 2 ba ab b 2 or 1

a b2 a 2 2ab b 2

Similarly, we obtain 2

a b2 a 2 2ab b 2

Thomson Brooks-Cole copyright 2007

EXAMPLE 2

(a) 2x 13x 5 6x 2 3x 10x 5 6x 2 7x 5 (b) x 62 x 2 12x 36 (c) 3x 14x 3 2x 6 34x 2 x 3 2x 12 12x 2 3x 9 2x 12 12x 2 5x 21

1

2 ■ REVIEW OF ALGEBRA

FRACTIONS

To add two fractions with the same denominator, we use the Distributive Law: a c 1 1 1 ac a c a c b b b b b b Thus, it is true that ac a c b b b But remember to avoid the following common error: a a a bc b c

|

(For instance, take a b c 1 to see the error.) To add two fractions with different denominators, we use a common denominator: a c ad bc b d bd We multiply such fractions as follows: a c ac b d bd In particular, it is true that a a a b b b To divide two fractions, we invert and multiply: a b a d ad c b c bc d


EXAMPLE 3

(a)

x3 x 3 3 1 x x x x

(b)

3 x 3x 2 xx 1 3x 6 x 2 x x 2 2x 6 2 2 x1 x2 x 1x 2 x x2 x x2

(c)

s2t ut s 2 t 2u s2t 2 u 2 2u 2

REVIEW OF ALGEBRA ■ 3

x xy 1 y y xy x xx y x 2 xy (d) y xy y xy yx y xy y 2 1 x x FACTORING

We have used the Distributive Law to expand certain algebraic expressions. We sometimes need to reverse this process (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest situation occurs when the expression has a common factor as follows: Expanding

3x(x-2)=3x@-6x Factoring

To factor a quadratic of the form x 2 bx c we note that x rx s x 2 r sx rs so we need to choose numbers r and s so that r s b and rs c. EXAMPLE 4 Factor x 2 5x 24. SOLUTION The two integers that add to give 5 and multiply to give 24 are 3 and 8. Therefore

x 2 5x 24 x 3x 8 EXAMPLE 5 Factor 2x 2 7x 4. SOLUTION Even though the coefficient of x 2 is not 1, we can still look for factors of the

form 2x r and x s, where rs 4. Experimentation reveals that 2x 2 7x 4 2x 1x 4 Some special quadratics can be factored by using Equations 1 or 2 (from right to left) or by using the formula for a difference of squares: 3

a 2 b 2 a ba b

The analogous formula for a difference of cubes is 4

a 3 b 3 a ba 2 ab b 2

which you can verify by expanding the right side. For a sum of cubes we have


5

a 3 b 3 a ba 2 ab b 2

EXAMPLE 6

(a) x 2 6x 9 x 32 (b) 4x 2 25 2x 52x 5 (c) x 3 8 x 2x 2 2x 4

(Equation 2; a x, b 3 ) (Equation 3; a 2x, b 5 ) (Equation 5; a x, b 2 )


EXAMPLE 7 Simplify

x 2 16 . x 2x 8 2

SOLUTION Factoring numerator and denominator, we have

x 2 16 x 4x 4 x4 x 2x 8 x 4x 2 x2 2

To factor polynomials of degree 3 or more, we sometimes use the following fact. 6 The Factor Theorem If P is a polynomial and Pb 0, then x b is a factor

of Px. EXAMPLE 8 Factor x 3 3x 2 10x 24. SOLUTION Let Px x 3 3x 2 10x 24. If Pb 0, where b is an integer, then b is

a factor of 24. Thus, the possibilities for b are 1, 2, 3, 4, 6, 8, 12, and 24. We find that P1 12, P1 30, P2 0. By the Factor Theorem, x 2 is a factor. Instead of substituting further, we use long division as follows: x 2 x 12 x 2 x 3 3x 2 10 x 24 x 3 2x 2 x 2 10x x 2 2x 12x 24 12x 24 Therefore

x 3 3x 2 10x 24 x 2x 2 x 12 x 2x 3x 4

COMPLETING THE SQUARE

Completing the square is a useful technique for graphing parabolas or integrating rational functions. Completing the square means rewriting a quadratic ax 2 bx c in the form ax p2 q and can be accomplished by: 1. Factoring the number a from the terms involving x. 2. Adding and subtracting the square of half the coefficient of x. In general, we have

ax 2 bx c a x 2 a x2


a x

b x c a

b x a b 2a

b 2a

2

2

c

2

b 2a

c

b2 4a

EXAMPLE 9 Rewrite x 2 x 1 by completing the square. 1

SOLUTION The square of half the coefficient of x is 4. Thus 2

x 2 x 1 x 2 x 14 14 1 (x 12 ) 34


EXAMPLE 10

2x 2 12x 11 2x 2 6x 11 2x 2 6x 9 9 11 2x 32 9 11 2x 32 7

QUADRATIC FORMULA

By completing the square as above we can obtain the following formula for the roots of a quadratic equation. 2 7 The Quadratic Formula The roots of the quadratic equation ax bx c 0 are

x

b sb 2 4ac 2a

EXAMPLE 11 Solve the equation 5x 2 3x 3 0. SOLUTION With a 5, b 3, c 3, the quadratic formula gives the solutions

3 s32 453 3 s69 25 10

x

The quantity b 2 4ac that appears in the quadratic formula is called the discriminant. There are three possibilities: 1. If b 2 4ac 0, the equation has two real roots. 2. If b 2 4ac 0, the roots are equal. 3. If b 2 4ac 0, the equation has no real root. (The roots are complex.)

These three cases correspond to the fact that the number of times the parabola y ax 2 bx c crosses the x-axis is 2, 1, or 0 (see Figure 1). In case (3) the quadratic ax 2 bx c can’t be factored and is called irreducible. y

0

y

x

0

y

x

0

x

FIGURE 1

Possible graphs of y=ax@+bx+c

(a) b@-4ac>0

(b) b@-4ac=0

(c) b@-4ac 0.

71. 3x2 + x − 6 is not irreducible because its discriminant is nonnegative: b2 − 4ac = 1 − 4(3)(−6) = 73 > 0.

72. The quadratic x2 + 3x + 6 is irreducible because b2 − 4ac = 32 − 4(1)(6) = −15 < 0. 73. Using the Binomial Theorem with k = 6 we have

6·5 4 2 6·5·4 3 3 6·5·4·3 2 4 a b + a b + a b + 6ab5 + b6 1·2 1·2·3 1·2·3·4 = a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6

(a + b)6 = a6 + 6a5 b +

74. Using the Binomial Theorem with k = 7 we have (a + b)7 = a7 + 7a6 b +

7·6 5 2 7·6·5 4 3 7·6·5·4 3 4 7·6·5·4·3 2 5 a b + a b + a b + a b + 7ab6 + b7 1·2 1·2·3 1·2·3·4 1·2·3·4·5

= a7 + 7a6 b + 21a5 b2 + 35a4 b3 + 35a3 b4 + 21a2 b5 + 7ab6 + b7 75. Using the Binomial Theorem with a = x2 , b = −1, k = 4 we have (x2 − 1)4 = [x2 + (−1)]4 = (x2 )4 + 4(x2 )3 (−1) + = x8 − 4x6 + 6x4 − 4x2 + 1

4·3 2 2 (x ) (−1)2 + 4(x2 )(−1)3 + (−1)4 1·2

76. Using the Binomial Theorem with a = 3, b = x2 , k = 5 we have (3 + x2 )5 = 35 + 5(3)4 (x2 )1 +

77. 78. 79. 80. 81. 82.

5·4 3 2 2 5·4·3 2 2 3 (3) (x ) + (3) (x ) + 5(3)(x2 )4 + (x2 )5 1·2 1·2·3

= 243 + 405x2 + 270x4 + 90x6 + 15x8 + x10 √ √ √ √ Using Equation 10, 32 2 = 32 · 2 = 64 = 8. u u √ √ 3 3 −2 −1 1 −1 3 −2 3 −1 √ √ = = =− = = 3 3 54 27 3 3 54 27 √ √ u √ 4 4 4 √ 32x4 32 x4 32 √ 4 √ √ x4 = 4 16 |x| = 2 |x|. Using Equation 10, 4 = = 4 4 2 2 2 s s √ s 3 xy x y = (xy)(x3 y) = x4 y 2 = x2 |y| √ √ √ √ √ Using Equation 10, 16a4 b3 = 16 a4 b3 = 4a2 b3/2 = 4a2 b b1/2 = 4a2 b b. u √ 5 6 √ 96a6 5 96a 5 √ = 32a5 = 2a = 5 3a 3a

83. Using Laws 3 and 1 of Exponents respectively, 310 × 98 = 310 × (32 )8 = 310 × 32 · 8 = 310 + 16 = 326 .

84. Using Laws 3 and 1, 216 × 410 × 166 = 216 × (22 )10 × (24 )6 = 216 × 220 × 224 = 260 . 85. Using Laws 4, 1, and 2 of Exponents respectively, 86. Using Laws 1 and 2,

x9 (24 )x4 16x9 + 4 x9 (2x)4 = = = 16x9 + 4 − 3 = 16x10 . 3 3 x x x3

an + 2n + 1 a3n + 1 an × a2n + 1 = = = a3n + 1−(n − 2) = a2n + 3 . an − 2 an − 2 an − 2

a2 a−3 b4 . 87. Using Law 2 of Exponents, −5 5 = a−3 − (−5) b4 − 5 = a2 b−1 = a b b 1 y+x (y + x)2 1 x−1 + y −1 + = (x + y) = = (x + y) 88. (x + y)−1 x y xy xy 89. By definitions 3 and 4 for exponents respectively, 3−1/2 =

√ √ √ √ √ 5 96 = 5 32 · 3 = 5 32 5 3 = 2 5 3 √ 2 91. Using definition 4 for exponents, 1252/3 = 3 125 = 52 = 25.

90. 961/5 = Thomson Brooks-Cole copyright 2007

1 1 = √ . 31/2 3

92. 64−4/3 =

1 1 1 1 = √ 4 = 4 = 3 4 256 644/3 64


93. (2x2 y 4 )3/2 = 23/2 (x2 )3/2 (y 4 )3/2 = 2 · 21/2

k√ l3 ks l3 √ √ x2 y 4 = 2 2 |x|3 (y 2 )3 = 2 2 |x|3 y 6

94. (x−5 y 3 z 10 )−3/5 = (x−5 )−3/5 (y 3 )−3/5 (z 10 )−3/5 = x15/5 y −9/5 z −30/5 = s 5 y 6 = y 6/5 by definition 4 for exponents. √ 3 96. ( 4 a ) = (a1/4 )3 = a3/4

x2

y 9/5 z 6

95.

1 1 1 97. √ 5 = 1/2 5 = 5/2 = t−5/2 (t ) t t √ 8 x5/8 1 x5 98. √ = 3/4 = x(5/8) − (3/4) = x−1/8 = 1/8 4 3 x x x u √ 1/2 1/2 1/2 1/4 1/4 1/2 t s t st 4 t 99. = = t(1/2) + (1/2) s(1/2) − (2/3) = (ts−1/6 )1/4 2/3 2/3 s s

t1/4 = t1/4 s(−1/6) · (1/4) = 1/24 s √ √ √ √ √ 4 2n + 1 4 −1 4 4 4 2n + 1 −1 100. r × r = r × r = r2n + 1 − 1 = r2n = (r2n )1/4 = r2n/4 = rn/2 √ √ √ (x − 9) 1 x−3 x−3 x+3 √ = = √ = ·√ 101. x−9 x−9 x+3 (x − 9) ( x + 3) x+3

1 1 1 1 1−x √ −1 √ −1 √ +1 −1 −1 −1 x x x x = = = √ x 102. = = · 1 x−1 x−1 x +x 1 1 1 √ +1 (x − 1) √ + 1 (x − 1) √ + 1 x √ +1 x x x x √ √ √ x3 − 64 x x−8 x x+8 x x−8 √ = = · √ 103. x−4 x−4 x x+8 (x − 4)(x x + 8) (x − 4)(x2 + 4x + 16) x2 + 4x + 16 √ √ [Equation 4 with a = x, b = 4] = (x − 4)(x x + 8) x x+8 √ √ √ √ √ √ 2 + h − (2 − h) 2+h+ 2−h 2+h+ 2−h 2+h− 2−h √ √ = √ = ·√ 104. h h 2+h− 2−h h 2+h− 2−h =

105. 106. 107. 108.

2 √ = √ 2+h− 2−h √ √ √ 2 3+ 5 2 3+ 5 2 3+ 5 √ = √ · √ = = 9−5 2 3− 5 3− 5 3+ 5 √ √ √ √ x+ y x+ y 1 1 √ √ = √ √ ·√ √ = x−y x− y x− y x+ y √ √ √ x2 + 3x + 4 − x2 3x + 4 x2 + 3x + 4 + x 2 2 = √ = √ x + 3x + 4 − x = x + 3x + 4 − x · √ x2 + 3x + 4 + x x2 + 3x + 4 + x x2 + 3x + 4 + x √ √ √ √ √ √ x2 + x − (x2 − x) x2 + x + x2 − x 2 2 2 2 √ √ x +x− x −x= x +x− x −x · √ = √ x2 + x + x2 − x x2 + x + x2 − x 2x √ = √ x2 + x + x2 − x

109. False. See Example 14(b).

110. False. See the warning after Equation 10. 16 a a 16 + a = + =1+ 16 16 16 16 1 1 1 xy = x+y = 6= x + y 112. False: −1 = 1 1 x + y −1 x+y + xy x y Thomson Brooks-Cole copyright 2007

111. True:

113. False. 114. False. See the warning on page 2.


115. False. Using Law 3 of Exponents, (x3 )4 = x3 · 4 = x12 6= x7 . 116. True.

117. |5 − 23| = |−18| = 18

118. |π − 2| = π − 2 because π − 2 > 0. √ √ √ √ 119. 5 − 5 = − 5 − 5 = 5 − 5 because 5 − 5 < 0. 120. |−2| − |−3| = |2 − 3| = |−1| = 1

121. If x < 2, x − 2 < 0, so |x − 2| = − (x − 2) = 2 − x.

122. If x > 2, x − 2 > 0, so |x − 2| = x − 2. + + x+1 if x + 1 ≥ 0 x+1 if x ≥ −1 123. |x + 1| = = −(x + 1) if x + 1 < 0 −x − 1 if x < −1 + + 2x − 1 if 2x − 1 ≥ 0 2x − 1 if x ≥ 12 124. |2x − 1| = = −(2x − 1) if 2x − 1 < 0 1 − 2x if x < 12 2 125. x + 1 = x2 + 1 (since x2 + 1 ≥ 0 for all x). t √ x2 > 12 126. Determine when 1 − 2x2 < 0 ⇔ 1 < 2x2 ⇔ x2 > 12 ⇔ + 1 − 2x2 if − √12 ≤ x ≤ √12 2 1 1 x < − √2 or x > √2 . Thus, 1 − 2x = 2x2 − 1 if x < − √12 or x > √12

⇔ |x| >

t

1 2

⇔

127. 2x + 7 > 3 ⇔ 2x > −4 ⇔ x > −2, so x ∈ (−2, ∞).

128. 4 − 3x ≥ 6 ⇔ −3x ≥ 2 ⇔ x ≤ − 23 , so x ∈ −∞, − 23 . 129. 1 − x ≤ 2 ⇔ −x ≤ 1 ⇔ x ≥ −1, so x ∈ [−1, ∞). 130. 1 + 5x > 5 − 3x ⇔ 8x > 4 ⇔ x > 12 , so x ∈

1 2

,∞ .

131. 0 ≤ 1 − x < 1 ⇔ −1 ≤ −x < 0 ⇔ 1 ≥ x > 0, so x ∈ (0, 1]. 132. 1 < 3x + 4 ≤ 16 ⇔ −3 < 3x ≤ 12 ⇔ −1 < x ≤ 4, so x ∈ (−1, 4]. 133. (x − 1)(x − 2) > 0.

Case 1: (both factors are positive, so their product is positive) x − 1 > 0 ⇔ x > 1, and x − 2 > 0 ⇔ x > 2, so x ∈ (2, ∞).

Case 2: (both factors are negative, so their product is positive)

x − 1 < 0 ⇔ x < 1, and x − 2 < 0 ⇔ x < 2, so x ∈ (−∞, 1).

Thus, the solution set is (−∞, 1) ∪ (2, ∞).


134. x2 < 2x + 8 ⇔ x2 − 2x − 8 < 0 ⇔ (x − 4)(x + 2) < 0. Case 1: x > 4 and x < −2, which is impossible. Case 2: x < 4 and x > −2. Thus, the solution set is (−2, 4). √ √ √ √ 135. x2 < 3 ⇔ x2 − 3 < 0 ⇔ x − 3 x + 3 < 0. Case 1: x > 3 and x < − 3, which is impossible. √ √ √ √ Case 2: x < 3 and x > − 3. Thus, the solution set is − 3, 3 . √ √ √ Another method: x2 < 3 ⇔ |x| < 3 ⇔ − 3 < x < 3.


√ √ √ √ √ 136. x2 ≥ 5 ⇔ x2 − 5 ≥ 0 ⇔ x − 5 x + 5 ≥ 0. Case 1: x ≥ 5 and x ≥ − 5, so x ∈ 5, ∞ . √ √ √ √ √ Case 2: x ≤ 5 and x ≤ − 5, so x ∈ −∞, − 5 . Thus, the solution set is −∞, − 5 ∪ 5, ∞ . √ √ √ Another method: x2 ≥ 5 ⇔ |x| ≥ 5 ⇔ x ≥ 5 or x ≤ − 5.

137. x3 − x2 ≤ 0 ⇔ x2 (x − 1) ≤ 0. Since x2 ≥ 0 for all x, the inequality is satisfied when x − 1 ≤ 0 ⇔ x ≤ 1. Thus, the solution set is (−∞, 1].

138. (x + 1)(x − 2)(x + 3) = 0 ⇔ x = −1, 2, or −3. Construct a chart: Interval

x+1

x−2

x+3

(x + 1)(x − 2)(x + 3)

x < −3

−

−

−

−

−1 < x < 2

+

−

+

−

−3 < x < −1 x>2

−

−

+

+

+

+

+

+

Thus, (x + 1)(x − 2)(x + 3) ≥ 0 on [−3, −1] and [2, ∞), and the solution set is [−3, −1] ∪ [2, ∞). 139. x3 > x ⇔ x3 − x > 0 ⇔ x x2 − 1 > 0 ⇔ x(x − 1)(x + 1) > 0. Construct a chart: Interval

x

x−1

x+1

x(x − 1)(x + 1)

x < −1

−

−

−

−

0 x when the last column is positive, the solution set is (−1, 0) ∪ (1, ∞).

140. x3 + 3x < 4x2

⇔ x3 − 4x2 + 3x < 0 ⇔ x x2 − 4x + 3 < 0 ⇔ x(x − 1)(x − 3) < 0. Interval

x

x−1

x−3

x(x − 1)(x − 3)

x 0. Then 1/x ≤ 1 ⇔ 1 ≤ x, and the solution set here is (−∞, 0) ∪ [1, ∞). Taking the intersection of the two solution sets gives the final solution set: −∞, − 13 ∪ [1, ∞).

143. C = 59 (F − 32) ⇒ F = 95 C + 32. So 50 ≤ F ≤ 95 ⇒ 50 ≤ 95 C + 32 ≤ 95 ⇒ 18 ≤ 95 C ≤ 63 ⇒ 10 ≤ C ≤ 35. So the interval is [10, 35].

144. Since 20 ≤ C ≤ 30 and C = 59 (F − 32), we have 20 ≤ 59 (F − 32) ≤ 30 ⇒ 36 ≤ F − 32 ≤ 54 ⇒ 68 ≤ F ≤ 86. So the interval is [68, 86].

145. (a) Let T represent the temperature in degrees Celsius and h the height in km. T = 20 when h = 0 and T decreases by 10◦ C for every km (1◦ C for each 100-m rise). Thus, T = 20 − 10h when 0 ≤ h ≤ 12. (b) From part (a), T = 20 − 10h ⇒ 10h = 20 − T ⇒ h = 2 − T /10. So 0 ≤ h ≤ 5 ⇒ 0 ≤ 2 − T /10 ≤ 5 ⇒ −2 ≤ −T /10 ≤ 3 ⇒ −20 ≤ −T ≤ 30 ⇒ 20 ≥ T ≥ −30 ⇒ −30 ≤ T ≤ 20. Thus, the range of temperatures (in ◦ C) to be expected is [−30, 20].

146. The ball will be at least 32 ft above the ground if h ≥ 32 ⇔ 128 + 16t − 16t2 ≥ 32 ⇔ 16t2 − 16t − 96 ≤ 0 ⇔ 16(t − 3)(t + 2) ≤ 0. t = 3 and t = −2 are endpoints of the interval we’re looking for,

and constructing a table gives −2 ≤ t ≤ 3. But t ≥ 0, so the ball will be at least 32 ft above the ground in the time

interval [0, 3].

147. |x + 3| = |2x + 1| ⇔ either x + 3 = 2x + 1 or x + 3 = − (2x + 1). In the first case, x = 2, and in the second case, x + 3 = −2x − 1

⇔

3x = −4 ⇔ x = − 43 . So the solutions are − 43 and 2.

148. |3x + 5| = 1 ⇔ either 3x + 5 = 1 or −1. In the first case, 3x = −4 ⇔ x = − 43 , and in the second case, 3x = −6 ⇔ x = −2. So the solutions are −2 and − 43 . 149. By Property 5 of absolute values, |x| < 3 ⇔ −3 < x < 3, so x ∈ (−3, 3). 150. By Properties 4 and 6 of absolute values, |x| ≥ 3 ⇔ x ≤ −3 or x ≥ 3, so x ∈ (−∞, −3] ∪ [3, ∞). 151. |x − 4| < 1 ⇔ −1 < x − 4 < 1 ⇔ 3 < x < 5, so x ∈ (3, 5). 152. |x − 6| < 0.1 ⇔ −0.1 < x − 6 < 0.1 ⇔ 5.9 < x < 6.1, so x ∈ (5.9, 6.1). 153. |x + 5| ≥ 2 ⇔ x + 5 ≥ 2 or x + 5 ≤ −2 ⇔ x ≥ −3 or x ≤ −7, so x ∈ (−∞, −7] ∪ [−3, ∞). 154. |x + 1| ≥ 3 ⇔ x + 1 ≥ 3 or x + 1 ≤ −3 ⇔ x ≥ 2 or x ≤ −4, so x ∈ (−∞, −4] ∪ [2, ∞). 155. |2x − 3| ≤ 0.4 ⇔ −0.4 ≤ 2x − 3 ≤ 0.4 ⇔ 2.6 ≤ 2x ≤ 3.4 ⇔ 1.3 ≤ x ≤ 1.7, so x ∈ [1.3, 1.7].

156. |5x − 2| < 6 ⇔ −6 < 5x − 2 < 6 ⇔ −4 < 5x < 8 ⇔ − 45 < x < 85 , so x ∈ − 45 , 85 . 157. a(bx − c) ≥ bc ⇔ bx − c ≥

bc a

⇔ bx ≥


158. ax + b < c ⇔ ax < c − b ⇔ x > 159. |ab| =

bc bc + ac +c = a a

⇔ x≥

bc + ac ab

c−b (since a < 0) a

√ √ √ s (ab)2 = a2 b2 = a2 b2 = |a| |b|

160. If 0 < a < b, then a · a < a · b and a · b < b · b [using Rule 3 of Inequalities]. So a2 < ab < b2 and hence a2 < b2 .