Math 140Quiz 1 - Summer 2004

Solution Review

Math 140Quiz 1 - Summer 2004

Solution Review

Math 140Quiz 1 - Summer 2004

Solution Review

(Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

Problem 1 (3)

Solve the equation: -7.3q + 1.9 = -59.7 – 1.7q.

-7.3q + 1.7q = -59.7 – 1.9

-5.6q = -61.6

q = 11

Thus, the answer is E).

xyxxyxy

yxy7274

2

282 224

4

254

Problem 2 (69)

Divide and simplify. Assume that all variables represent positive real numbers.

4

65

2

56

y

yx

xyxy

yxy74

2

282 24

4

254

4

254

2

282

y

yxy

Problem 3 (38)

Simplify the radicals and combine any like terms. Assume all variables represent positive real numbers.

19•213 - 3•(2 • 27)13 =

19•213 - 3•213(33)13 =

19•213 - 9•213 =

Problem 4 (79)

Perform the indicated operation and

simplify: 4 + 2w .

w - 2 2 - w

_______________

__________

(101/2)2 - 32

Problem 5 (55)

Rationalize the denominator: 10 .

Problem 6 (55)

Use rational exponents to simplify the radical. Assume that all variables represent positive numbers.

(9•9)112 = (34)12 =

342 = 31=

Problem 7 (52)

Solve the equation: .

Alternate approach: Multiply by LCD = 12x. Then,

12x – 7(12x)/(3x) = (10/4)(12x)

12x – 28 = 30x

-18x = 28

x = -28/18 = -14/9

Problem 7 ContinuedSolve the equation: .

Problem 8 (66)

Solve the equation by a u-substitution and factoring.

x4 + x2 – 2 = 0

Let u = x2 . Then the equation is

u2 + u – 2 = 0

There are only two factoring possibilities:

(u - 1) (u + 2) & (u + 1) (u - 2).

But only the combination (u + 2) (u - 1) works.

u2 + u – 2 = (u + 2) (u - 1) = 0 => u = 1 or -2.

Since u= x2 > 0, drop –2 case & deduce x2 = u = 1.

Hence, x2 – 1= (x + 1) (x - 1) = 0 => x = -1 or 1.

Problem 9 (62)

Use radical notation to write the expression. Simplify if possible: .

Note: (-1) = (-1)3 & 512 = 29. Thus,

-512 x12 = (-1)3 29 x12 =>

(-512 x12 )1/3 = [(-1)3 29 x12 ]1/3 =

(-1)3/3 29/3 x12/3 = - 23 x4 = - 8x4

Problem 10 (41)

A rectangular carpet has a perimeter of 236 inches. The length of the carpet is 94 inches more than the width. What are the dimensions of the carpet?

Let W = width & L = length = W + 94

Perimeter = 2L + 2W = 236

2(W + 94) + 2W = 236

4W=236-188=48 => W=12" & L=106"

Problem 11 (38)

Solve by completing the square: x2 + 8x = 3.

x2 + 8x + (8/2)2 = 3 + (8/2)2

x2 + 8x + 16 = (x + 4)2 = 19

x + 4 = 191/2 or x + 4 = -191/2

x = -4 + 191/2 or x = -4 - 191/2

{-4± }

Problem 12 (31)

Solve the equation: 18n2 + 78n = 0.

18n2 + 78n = 0

6n(3n + 13) = 0

n = 0 or (3n + 13) = 0

n = 0 or n = -13/3

{-13/3, 0}

Problem 13 (34)

Solve the equation by factoring: x3 + 6x2 - 7x = 0.

x(x2 + 6x - 7) = 0

x(x + 7) (x - 1) = 0

x = 0 or x + 7 = 0 or x - 1 = 0

x = 0 or x = - 7 or x = 1

{-7, 0, 1}

Problem 14 (69) The manager of a coffee shop has one type of

coffee that sells for $10 per pound (lb) and another type that sells for $15/lb. The manager wishes to mix 40 lbs of the $15 coffee to get a mixture that will sell for $14/lb. How many lbs of the $10 coffee should be used?

Let t = amt of $10/lb & f = amt of $15/lb = 40

To have value equal: 10t +15f = 14(40+t).

10t +15(40) = 560+ 14t or 10t +600 – 14t = 560

-4t = -40 => t = 10 pounds

5 6 7 8 9 10

Problem 15 (38)

Write each expression in interval notation.

Graph each interval. x > 6

Recall rules: + => Open(+) right/left(-) end

Note: x > 6 => 6 < x so only left end is determined.

Open end Closed end

Left side: a < x => (a, a < x => [a,

(6, ............................) 5 6 7 8 9 10

(

Problem 16 (21)

Write each expression in interval notation.

Graph each interval. -2 < x < 1

Recall notation rules:

Open end Closed end

Left side: a < x => (a, a < x => [a,

Right side: x < a => , a) x < a => , a]

(-2, 1]

-3 -2 -1 0 1 2 -3 -2 -1 0 1 2( ]

Problem 17 (48)

Solve the equation.

p2 - 5p + 81 = (p + 4)2 =

p2 + 8p + 16

-13p + 65 = 0

-13p = -65

p = 5 ________________________ {5} is solution set.

Always check when squaring radical equations since spurious roots can be introduced. Check:

48152 ppp

?4581)5(55 Does 2

YES!

Problem 18 (17)

Solve the equation: |5m + 4| + 8 = 10 .

5m + 4 = 2 or 5m + 4 = -2

5m = -2 or 5m = -6

m = -2/5 or m = -6/5

{-2/5, -6/5}

)47(

)47)(47(

xy

xy

xy

xyxy

Problem 19 (45)

Simplify the complex fraction.

122 )( 471649

yxxy

xy

xy

xyxy )47)(47(

xy 47

Problem 20 (41)

Solve the inequality. Write answer in interval notation.

|r + 4| > 2

r + 4 > 2 or r + 4 < -2

r > 2 - 4 or r < -2 - 4

r > - 2 or r < -6

(-, -6] or [-2, )