Excel Review Center Solution to Take Home Exam – Algebra 1

Question 1

A. the final digit whether zero or not zero Question 2

C. leading digit Question 3

D. 5 Question 4

D. 7 Question 5

A. 3 Question 6

B. 0.01414 Question 7

C. evolution Question 8

D. involution Problem 9

x 1 2x47 2x

3 4

4x 4 6x47 2x

12

4x 4 6x 564 24x

4x 6x 24x 564 4

34x 560

x 16.47

++ = −

+ += −

+ + = −

+ + = −

=

=

CALCULATOR: Input the equation then use the SHIFT CALC function: Answer: x = 16.47 Problem 10

( ) ( ) ( )

( )( )( )

( )( )( )

( )( )( )

( )( )

( )( )( )

2 2 2

2 2

2 2 2

2

5x x 3 2x 1

2x 7x 3 2x 3x 2 x x 6

5x x+3 2x+1= - +

(2x+1)(x+3) (2x+1)(x-2) (x-2)(x+3)

5x x 2 x 3 2x 1

2x 1 x 3 x 2

5x 10x x 6x 9 4x 4x 1

2x 1 x 3 x 2

8x 12x 8

2x 1 x 3 x 2

4 2x 1 x 2

2x 1 x 3 x 2

4

x 3

+ +− +

+ + − − + −

− − + + +=

+ + −

− − − − + + +=

+ + −

− −=

+ + −

+ −=

+ + −

=+

CALCULATOR: Input the equation then use the CALC function. Let x = 1:

( )

( ) ( ) ( ) ( )

( )

( ) ( )

2 2 2

2 2 2

5x x 3 2x 1

2x 7x 3 2x 3x 2 x x 6

5 1 2 1 11 3

2 1 7 1 3 2 1 3 1 2 1 1 6

1

+ +− +

+ + − − + −

++− +

+ + − − + −

=

Then from the choices, Let x = 1:

4 4

x+3 1+3

1 satisfies the given equation

→

→

Problem 11

( ) ( ) ( )

( )

( ) ( )

n 1 2n 1

2 n 1

n 2n 2

xy a b a b

xy a b

xy a b

− −

−

−

− −

−

−=

( ) ( )n 1 2n 2

xy a b− −

−

xy=

CALCULATOR: Assume values for variables a,b, x and y, then use the CALC function.(Same with the previous problem.) Problem 12

( )( )( )( )

( )( )( )( )

2 2 2

2 2

2

a b a a 2b

ab a ab 2b a b

a b a b a a 2b a

ab a 2b a b a b b

− −⋅ ⋅

− − −

− + −= =

− + −

CALCULATOR: (Same with the previous problem) Input the equation then use the CALC function. Let a=1, b=2:

( ) ( )

( )( )

( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

2 2 2

2 2

2 2 2

2 2

a b a a 2b

ab a ab 2b a b

1 2 1 1 2 2

1 2 1 21 1 2 2 2

1

2

− −⋅ ⋅

− − −

− −= ⋅ ⋅

−− −

=

Then from the choices, Let a = 1, b= 2:

a 1satisfies the given equation

b 2= →

Problem 13

( )( )

2

2x 3

x x 2 x 2

2x 3

x 2 x 1 x 2

2x 3x 3

x 3

=+ − +

=+ − +

= −

=

CALCULATOR: Input the equation then use the SHIFT CALC function:

Answer: x = 3 Problem 14

2

2

4 3 1 6

x 1 x 1 2x x 1

4x 4 3x 3

x 1

+ = −− + −

+ + −

− ( )

2

2

x 1 12x

2x x 1

− −=

−

2 2 2

2

8x 8x 6x 6x x 1 12x

13x 14x 1 0

1x , x 1

13

+ + − = − −

+ + =

= − = −

CALCULATOR: Input the equation then use the SHIFT CALC function: Answer: x = -1/13 or x = -1

Problem 15

( )( )

( )( )( )

4 2 2 2

2

x 5x 36 x 9 x 4

x 9 x 2 x 2

+ − = + −

= + − +

Problem 16

( ) ( )

( )( )

( )( )( )

3 2 2

2

x 3x 4x 12 x x 3 4 x 3

x 3 x 4

x 3 x 2 x 2

+ − − = + − +

= + −

= + − +

Problem 17

( )( )

( )( )( )

4 2 2

2

81 x 9 x 9 x

9 x 3 x 3 x

− = + −

= + − +

Problem 18

( )

( ) ( ) ( )

( )( )

2 2

3x y 2y 1

x y x x y x y

3x y x 2y x y x x y

x x y x y

+− −

− − +

+ − + − −=

− +

( )( )

( )( )

( )

( )

2 2 2

2 22 2

2 2

3x xy 2xy 2y x xy

x x y x y

2 x y2x 2y 2

x x y x y xx x y

+ − − − +=

− +

−−= = =

− + −

Problem 19

( ) ( )

( )( )

( ) ( )

( )( )

( ) ( )

( )( )

2 2

2 2

x 1 x 2x 1 x 2

x 2 x 1x 2 x 11 1 x 2 x 1

x 1 x 2 x 2 x 1

x 1 x 2

x 2 x 1

− − −− −

− − −− − =− − −

−− − − −

− − −=

− −

( )( )x 2 x 1 − −

( ) ( )x 2 x 1

− − −

Excel Review Center Solution to Take Home Exam – Algebra 1

( ) ( )

( ) ( )

2 22 2x 1 x 2 x 2x 1 x 4x 4

x 2 x 1 1

3 2x

− − − − + − + −= =

− − − −

= −

Problem 20

( )3 1 3

1 43 2

2 4

5x y 1 xyx y

15x y 3 3

−− − −−

−= =

Problem 21

( ) ( )3311 2 3 2 2 3 2 23 3316x y 2 2 x x y 2x 2x y= =

Problem 22

2 2

2

2x 1 x 8

x 5 7x

14x 7x x 3x 40

15x 10x 40 0

4x 2,

3

+ +=

− −

− − = + −

+ − =

= −

Problem 23

( ) ( )

( ) ( )( ) ( )

( ) ( )( )

( ) ( )( )

( ) ( )

2

22

2

2 2

2 2

2

x 1 2x 3 8x 1 0

x 1 8x 1 2x 3

x 1 8x 1 2 8x 1 2x 3 2x 3

x+1-8x-1-2x-3 = -2 8x+1 2x+3

3 3x 1 4 8x 1 2x 3

9 9x 6x 1 4 16x 26x 3

81x 54x 9 64x 104x 12

17x 50x 3 0

1x 3, x discarded

17

Substitute the tw

+ + + − + =

+ = + − +

+ = + − + + + +

− + = + +

+ + = + +

+ + = + +

− − =

= = − →

o values in the

equation given, the value that will

satisfy the equation

is x = 3.

Problem 24

The LCM is the product of the highest power of each prime factor of the given numbers

26 13 2

39 13 3

66 11 3 2

least common multiple is:

=13 2 3 11 858

= ×

= ×

= × ×

∴

× × × =

Problem 25

2

2

2

2 2 2

12 2 3

18 3 2

21 3 7

25 5

35 3 7

LCM is: 2 3 7 5 6300

= ×

= ×

= ×

=

= ×

∴ = × × × =

Problem 26

15 1 15

28 1 2 2 7

Greatest common divisor is 1

= ×

= × × ×

∴

Problem 27

12 3 4

16 4 4

Greatest common divisor is 4

= ×

= ×

∴

Question 28

A. imaginary Problem 29

33 3 2

5 2 5 2 5 2

35 22

a a alog log

c b c b c b

loga logc logb

3loga 5logc 2logb

2

= =

= − −

= − −

Problem 30

( )( )( )

10

10

10 10

10 10 10

10

log 3 0.4771

log 4 0.6021

log 12 log 3 4

log 12 log 3 log 4

log 12 1.0792

=

=

=

= +

=

Problem 31

4

14 4

4

log 7 n

1log log 7

7

log 7 n

−

=

=

− = −

Problem 32

( )2

3

2

log x - 8x 2

x 8x 9

(x 9)(x 1) 0

x 9 l x 1

=

− =

− + =

= = −

Problem 33

b b

b b

b

2x

2x

log y 2x log x

log y log x 2x

ylog 2x

x

yb

x

y xb

= +

− =

=

=

=

Problem 34

( )

2 3 1

2 3

2 3

2 3

2 3

2 3

x x x

x x x

x x x

x x x

a c b

a c b b

log a c b logb

loga logc logb logb

xloga xlogc xlogb logb

logbx

loga logc logb

− +

− −

− −

− −

=

=

=

+ + =

− − =

=− −

Problem 35

( ) ( )

( ) ( )

( )

2

2

2log 3 x log2 log 22 2x

log 3 x log2 22 2x

3 x 44 4x

By quadratic formula :

x 7 x 5

Substitute the values in the given

equation, x=7 will five log of negative

number, and x=-5 will result to a

logarithm of a pos

− = + −

− = −

− = −

= = −

itive value.

Answer is x 5∴ = −

Problem 36

3

2

2 6.278

6log2x log 6.278

x

log12x 6.278

12x 10

x 397.56

+ =

=

=

=

Problem 37

2 2

2

log 2 log x 2

log 2x 2

2x 4

x 2

+ =

=

=

=

Problem 38

3 log x

2

x 100x

3logx(logx) log(100x)

3logx(logx) log100 logx

3(logx) logx 2 0

=

=

= +

− − =

Let y = logx 2

2

3(logx) logx 2 0

3y y 2 0

y 1,y 2/3(absurd)

When,logx 1

x 10

− − =

− − =

= = −

=

=

Problem 39

( )4 3log log 5 0.275 (use calculator)= →

Problem 40

xylne xy=

Excel Review Center Solution to Take Home Exam – Algebra 1 Problem 41

( )

3 3 3

3 3

2

2

log 4x log x log 144

log 4x x log 144

4x 144

x 36

x 6 discarded

x 6 Answer

+ =

× =

=

=

= − →

= →

Problem 42

( )

5 5 5

3 3

2

2

log 4x log x log 100

log 4x x log 100

4x 100

x 25

x 5

+ =

× =

=

=

=

Question 43

D. positive value or zero Problem 44

From sum of roots formula:

b 10Sum 2

a 5

− = − = − =

Problem 45

2 2

2 2

2 2

2 2

a x 4c x 10c 5a 4acx

a x 4c x 4acx 5a 10c

x(a 4c 4ac) 5a 10c

5(a 2c)x

(a 4c 4ac)

5(a 2c)x

(a 2c)(a 2c)

5x

(a 2c)

+ − = −

+ + = +

+ + = +

+=

+ +

+=

+ +

=+

Problem 46

( ) ( )

( )

( )

1 2 1 2

2

2

5x 4 2x 1 1

5x 4 2x 1 2 2x 1 1

5x 2x 4 2 2 2x 1

3x 6 2 2x 1

9x 36x 36 4 2x 1

9x 44x 32 0

x 4

8x discarded

9

Substituting the two values in

the given equation, only x = 4

will satisfy the equation.

− = + +

− = + + + +

− − − = +

− = +

− + = +

− + =

=

= →

Problem 47

( )32

Remainder,R:

R f(4 3) 6(4 3) 30 9 4 3 2= = − + =

Problem 48

( ) ( ) ( ) ( )4 3 2

Remainder,R:

R f(1 3)

R 3 1 3 5 1 3 5 1 3 10 1 3 1

R 2

=

= + − + −

=

Problem 49

Using Quadratic Formula: x 1 2i= − ± Problem 50

( ) ( ) ( ) ( ) ( )6 5 4 2

Remainder,R:

R f( 5)

R -5 +7 -5 +10 -5 - -5 - -5

R 0

= −

=

=

Problem 51

( ) ( ) ( )4 3 2

Remainder,R:

f(2) 16

2 a 2 5 2 b(2) 6 16

16 8a 20 2b 6 16

8a 2b 26 0

4a b 13 0

=

+ + + + =

+ + + + =

+ + =

+ + = →�

When divided by x+1:

( ) ( ) ( )4 3 2

Remainder,R:

f( 1) 10

1 a 1 5 1 b( 1) 6 10

1 a 5 b 6 10

a b 2 0

− =

− + − + − + − + =

− + − + =

+ − = →�

Solving the two equations simultaneously:

4a b 13 0

a b 2 0

+ + = →

+ − = →

�

�

We get, a = -5 and b = 7. Answer: a = -5 Problem 52

Solving for f(k)=k.

2

2

(k 3)(k 4) 4 k

k 4k 3k 12 4 k

k 2k 8 0

(k 4)(k 2) 0

k 4 l k 2

+ − + =

− + − + =

− − =

− + =

= = −

Problem 53

Using long division:

2

3 2 5 4 3 2

5 4 3 2

4 3 2

4 3 2

3 2

3 2

3x 6x 8

x 2x 6 3x 0x 4x 2x 36x 48

3x 6x 0x 18x

0 6x 4x 16x 36x

0 6x 12x 0x 36x

8x 16x 48

8x 16x 48

0

+ +

− + + − + + +

− + +

+ − − +

+ − + +

− +

− +

Problem 54

Solving for the roots of the given equation by Q.F.:

x 5 / 2 l x 1= = −

Taking the reciprocal, 1 2x ' 2 5 l x ' 1= = −

Thus, the required equation is:

2

2

2

2(x )(x 1 0

5

2x 2x x 0

5 5

5x 5x 2x 2 0

5x 3x 2 0

− + =

+ − − =

+ − − =

+ − =

Problem 55

2

2

2

1 2(x )(x ) 0

2 3

2x x 1x 0

3 2 3

x 1x 0

6 3

6x x 2 0

− + =

+ − − =

+ − =

+ − =

Problem 56

Solving for the roots of the given equation by Q.F., we get

( )( )

2

2

b b 4acx

2a

7 7 4 1 2x

2(1)

7 57 7 57x l x

2 2

− ± −=

− ± − −=

− + − −= =

The new roots are:

7 57 7 57x l x

2 2

− += =

Thus, the required equation is: 2x 7x 2 0− − =

Problem 57

Let, x1 be one of the root. 2x1 be the other root Then, from sum of roots formula:

( )1 1

1

x 2x k

3x k

+ = − −

= →�

Also, from product of roots formula:

( )

( )

( )

1 1

2

1

2

1

1

x 2x 18

2 x 18

x 9

x 3

=

=

=

= ±

Thus solving for k from equation 1 we get k = 3x1 = 3(±3) k = ±9 Problem 58

Express the given in general form:

24x 8x 5 0− + =

By inspection, we get:

Excel Review Center Solution to Take Home Exam – Algebra 1 A = 4, B = -8, and C = 5 Solving for discriminant:

( ) ( )( )

2

2

D B 4AC

D 8 4 4 5

D 16

= −

= − −

= −

Problem 59

( )( )

2

2

2

Using the discriminant, one real solution:

D B 4AC 0

D k 4 4 1 0

k 16

k 4

= − =

= − =

=

=

Problem 60

Substitute x = 3 to the given equation.

( ) ( )3 2

3 2 3 3k 0

k 3

− − =

=

Problem 61

From sum of roots formula:

1 2

1 2

kx x

2

kx x

2

− + = −

+ = →�

From the given that the difference between roots is 5/2.

1 2

5x x

2− = →�

From product of roots formula:

( )( )1 2

3kx x

2= →�

Manipulating the 3 equations:

k = - 1, k = 25

Problem 62

If one root is the reciprocal of the other, then

the product of the roots must be 1.

C1

A

5k1

2

2k

5

=

=

=

Problem 63

Using discriminant:

( ) ( )( )

2

2

D (B) 4AC

D 12 4 4 9

D 0

= −

= − −

=

Since the discriminant is zero, the roots

are “rational and equal”.

Problem 64

Let y = x1/5 22y 5y 3 0+ − =

Solving for y using Q.F., we get:

1 2

1y l y 3

2= = −

Solving for x:

1 5 1 51x , x 3

2

1x , x 243

32

= = −

= = −

Problem 65

If the roots are equal, the discriminant must be zero:

2

2

2

D (2k 4) 4(9k) 0

4k 16k 16 36k 0

4k 20k 16 0

k 1, k 4

= + − =

+ + − =

− + =

= =

Problem 66

(x y)log5 2

2x y

log5

(2x y)log2 1

12x y

log2

+ =

+ = →

− =

− = →

�

�

Adding 1 and 2:

2 13x

log5 log2

x 2.06

= +

=

Problem 67

Let y = x2 2y 10y 9 0

y 1,y 9

− + =

= =

But, y =x2 2 2x 1, x 9

x 1, x 3

= =

= ± = ±

Problem 68

( )( )

3 2

7x 2 x 3 x

5

5x 2x 37x 42 0

− + −

= − − + =

Problem 69

x 3y 4z 15

2x 4y 5z 12

3x y 6z 29

+ + =

− + + =

+ + =

Using calculator, we get: x=2

Problem 70

x 2y z 10

2x y 2z 3

3x 2y 3z 6

+ − =

− + = −

+ + =

Using calculator, we get: x=2

Problem 71

From the formula:

( ) ( )

( )

n r 1 r 1

r 1

7 4 1 4 1

4 1

4 3

4 3

4th term nC x y

4th term 7C a 2x

35a 8x

280a x

− + −

−

− + −

−

=

= −

=

= −

Problem 72

From the formula:

( ) ( )

( ) ( )

n r 1 r 1

r 1

8 5 1 5 1

5 1

4 5 1

4 4

5th term nC x y

5th term 8C 3y 4w

70 3y 4w

1,451,520y w

− + −

−

− + −

−

−

=

= −

= −

=

Problem 73

From the formula:

( )

rth n r r

r

55 2 5

5

10 5

y term nC x y

y 10C 2x y

8064x y

−=

=

=

Problem 74

From the formula:

( ) ( )

th n r 1 r 1

r 1

9 r 1 r 12 1

r term nC x y

2x x

− + −

−

− + −−

=

=

Collect variables:

( ) ( )9 r 1 r 1

2 1 0

20 2r 1 r 0

21 3r 0

x x x

x x x

x x

21 3r 0

r 7

− + −−

− −

−

=

=

=

− =

=

Solving for the coefficient of the 7th term:

( )

th 2 9 7 1 r 1

7 1

3

6

17 term 9C (2x ) ( )

x

9C 2

672

− + −

−=

=

=

Problem 75

From the formula:

( )

n r 1 r 1

r 1

16 6 15

6 1

11

11

6th term nC x y

16th term 16C 3

2a

14368 ( 243)

2048a

66339

128a

− + −

−

− +

−

=

= −

= −

= −