Assignment 8

course MTH158

6/19/09 1:00 A.M. This was a challenging assignment for me. I can look back through the book and find the formulas and do the problems. Not sure how well I will do when I dont have the book to refer to on tests.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

008. `* 8

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Question: * R.8.12. Simplify the cube root of 54

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Your solution:

First factor the 54 so one number is a cube

cube root of 27.2

=3*cube root of 2

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * The cube root of 54 is expressed as 54^(1/3).

The number 54 factors into 2 * 3 * 3 * 3, i.e., 2 * 3^3. Thus

54^(1/3) = (2 * 3^3) ^(1/3)

= 2^(1/3) * (3^3)^(1/3)

= 2^(1/3) * 3^(3 * 1/3)

= 2^(1/3) * 3^1

= 3 * 2^(1/3), i.e.,

3 * cube root of 2.

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ).

3x* (3)^1/3

Confidence Assessment: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

The cube root of (3 x y^2 / (81 x^4 y^2) ) is

(3 x y^2 / (81 x^4 y^2) ) ^ (1/3) =

(1 / (27 x^3) ) ^(1/3) =

1 / ( (27)^(1/3) * ^x^3^(1/3) ) =

1 / ( (3^3)^(1/3) * (x^3)^(1/3) ) =

1 / ( 3^(3 * 1/3) * x^(3 * 1/3) ) =

1 / (3 * x) =

1 / (3x).

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Self-critique (if necessary):

I set up the problem and tried to do it alot different than you did. The way you showed here was alot simpler

than in the book. I will use these to study by.

Self-critique Rating:2

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Question: * R.8.30. Simplify 2 sqrt(12) - 3 sqrt(27).

2*sqrt(4*3)-3sqrt(9*3)

=4 sqrt(3) - 9 sqrt(3)

=4-9*sqrt(3)

Confidence Assessment: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

2 sqrt(12) - 3 sqrt(27)

= 2 sqrt( 2*2*3) - 3 sqrt(3*3*3)

= 2 sqrt(2^2 * 3) - 3 sqrt(3^3)

= 2 sqrt(2^2) sqrt^3) - 3 sqrt(3^2) sqrt(3)

= 2 * 2 - 3 * 3 sqrt(3)

= 4 - 9 sqrt(3).

Extra Question: What is the simplified form of (2 sqrt(6) + 3) ( 3 sqrt(6)) and how did you get this result?

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Your solution:

Distribute through the parentheses

2sqrt6*3sqrt6 + 9sqrt6

6 sqrt (6)^2 + 9 sqrt 6

6 * 6 which is the sqrt of (6)^2 + 9 sqrt 6

36 + 9 sqrt 6

Confidence Assessment: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

(2*sqrt(6) +3)(3*sqrt(6)) expands by the Distributive Law to give

(2*sqrt(6) * 3sqrt(6) + 3*3sqrt(6)), which we rewrite as

(2*3)(sqrt6*sqrt6) + 9 sqrt(6) =

(6*6) + 9sqrt(6) =

36 +9sqrt(6).

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8. Expand (sqrt(x) + sqrt(5) )^2

sqrt(x) * sqrt(x) + sqrt(x) * sqrt (5) + sqrt (5) * sqrt(x) + sqrt (5) * sqrt (5) (distributive property)

sqrt(x^2) + sqrt (x) * sqrt (5) + sqrt (5) * sqrt (x) + sqrt (25)

x + sqrt (x) * sqrt (5) + sqrt (5) * sqrt (x) + 5

x + 2 *sqrt (5X) + 5

Confidence Assessment: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

(sqrt(x) + sqrt(5) )^2

= (sqrt(x) + sqrt(5) ) * (sqrt(x) + sqrt(5) )

= sqrt(x) * (sqrt(x) + sqrt(5) ) + sqrt(5) * (sqrt(x) + sqrt(5) )

= sqrt(x) * sqrt(x) + sqrt(x) * sqrt(5) + sqrt(5) * sqrt(x) + sqrt(5) * sqrt(5)

= x + 2 sqrt(x) sqrt(5) + 5.

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Self-critique (if necessary): I worked this problem over and over again

until I came up with the correct answer. This really threw me. Im not sure why?

You didn't show your earlier trials (you're not expected to do so, though you may) so I can't guess where you were originally going wrong, but you did fine in the end.

Self-critique Rating:

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Question:

* R.8.42. What do you get when you rationalize the denominator of 3 / sqrt(2) and what steps did you follow to get this result?

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Your solution:

multiplied the numerator and denominator by sqrt of (2)

3 * sqrt(2)/(sqrt(2)*sqrt(2))

3*sqrt(2)/(sqrt(4)

=3*sqrt(2)/2

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

Starting with 3/sqrt(2) we multiply numerator and denominator by sqrt(2) to get

(2*sqrt(2))/(sqrt(2)*sqrt(2)) =

(3 sqrt(2) ) /2.

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8.48. Rationalize denominator of sqrt(2) / (sqrt(7) + 2)

Multiply numerator and denominator by sqrt(7) -2

sqrt(2)*sqrt(7) - (2)/(sqrt(49)-4)

=sqrt(2) * sqrt(7)-2/7 - 4

=sqrt(2) * sqrt(7)-2/3

To rationalize the denominator sqrt(7) + 2 we multiply both numerator and denominator by sqrt(7) - 2.

We obtain

( sqrt(2) / (sqrt(7) + 2) ) * (sqrt(7) - 2) / (sqrt(7) - 2)

= sqrt(2) * (sqrt(7) - 2) / ( (sqrt(7) + 2) * ( sqrt(7) - 2) )

= sqrt(2) * (sqrt(7) - 2) / (sqrt(7) * sqrt(7) - 4)

= sqrt(2) * (sqrt(7) - 2 ) / (7 - 4)

= sqrt(2) * (sqrt(7) - 2 ) / 3.

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Self-critique (if necessary):

Self-critique Rating:

Extra Question: What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?

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Your solution:

multiplied 3/1*1/6

to come up with x^3/6 = x ^1/2

Confidence Assessment: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * Express radicals as exponents and use the laws of exponents.

(x^3)^(1/6) =

x^(3 * 1/6) =

x^(1/2). **

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8.60. Simplify 25^(3/2).

=125

sqrt(25)^3

5^3=125

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

25^(3/2) =

(5^2)^(3/2) =

5^(2 * 3/2) =

5^(2 * 3/2) =

5^3.

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8.72. Simplify and express with only positive exponents:

(xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4).

(x^1/4 y^1/4)(xy)/((x^6/4 y^3/4))

x^1/4-6/4) (y^1/4-3/4)

= xy/(x^5/4 y^1/2)

=x^1/1-5/4)=-1/4

y^1/1-1/2)=1/2

=y^1/2/(x^1/4)

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

(xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4)

= x^(1/4) * y^(1/4) * (x^2)^(1/2) * y^2 ^ (1/2) / ( (x^2)^(3/4) * y^(3/4) )

= x^(1/4) * y^(1/4) * x^(2 * 1/2) * y^(2 * 1/2) / ( (x^(2 * 3/4) * y^(3/4) )

= x^(1/4) y^(1/4) * x^1 * y^1 / (x^(3/2) y^(3/4) )

= x^(1 + 1/4) y^(1 + 1/4) / (x^(3/2) y^(3/4) )

= x^(5/4) y^(5/4) / (x^(3/2) y^(3/4) )

= x^(5/4 - 3/2) y^(5/4 - 3/4)

= x^(5/4 - 6/4) y^(2/4)

= x^(-1/4) y^(1/2)

= y^(1/2) / x^(1/4).

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Self-critique (if necessary):

Self-critique Rating:

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Question: * R.8.84. Express with positive exponents:

( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2).

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Your Solution:

((3-x + x^2(1/3 - 1/x))/9 -x^2

= (1-x+x^2(1/3-1/x))/3-x^2

Confidence Assessment: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2)

=

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Self-critique (if necessary):

Could you explain this problem...I have no idea what I done. I confused myself more than I helped.

Self-critique Rating:1

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Question: * R.8.108. v = sqrt(64 h + v0^2); find v for init vel 0 height 4 ft; for init vel 0 and ht 16 ft; for init vel 4 ft / s and height 2 ft.

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Your Solution:

sqrt (256+0) = 16'/s

sqrt (1024+0) = 32'/s

sqrt (128+4^2) = (128+16) = sqrt(144) = 12'/s

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

If initial velocity is 0 and height is 4 ft then we substitute v0 = 0 and h = 4 to obtain

v = sqrt(64 * 4 + 0^2) = sqrt(256) =16.

If initial velocity is 0 and height is 16 ft then we substitute v0 = 0 and h = 4 to obtain

v = sqrt(64 * 16 + 0^2) = sqrt(1024) = 32.

Note that 4 times the height results in only double the velocity.

If initial velocity is 4 ft / s and height is 2 ft then we substitute v0 = 4 and h = 2 to obtain

v = sqrt(64 * 2 + 4^2) = sqrt(144) =12.

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Self-critique (if necessary):

Self-critique Rating:

Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result?

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Your solution:

Same as the cube root of 24 ^1 Factor 24 to 8*3 which 2 is the cube root of 8. Move the 2 outside the radicand and leave

the 3 inside

=2 * cuberoot(3)

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * (24)^(1/3) =

(8 * 3)^(1/3) =

8^(1/3) * 3^(1/3) =

2 * 3^(1/3) **

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Self-critique (if necessary):

Self-critique Rating:

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Question:

Extra Question: What is the simplified form of (x^2 y)^(1/3) * (125 x^3)^(1/3) / ( 8 x^3 y^4)^(1/3) and how did you get this result?

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Your solution:

((x^2/3) (y^1/3) (5) (x))/2 (x) (y^4/3)

x cancels (y^(1/3 - 4/3) = -y

5(x^2/3)/2y

-

Confidence Assessment: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * (x^2y)^(1/3) * (125x^3)^(1/3)/ ( 8 x^3y^4)^(1/3)

(x^(2/3)y^(1/3)* (5x)/[ 8^(1/3) * xy(y^(1/3)]

(x^(2/3)(5x) / ( 2 xy)

5( x^(5/3)) / ( 2 xy)

5x(x^(2/3)) / ( 2 xy)

5 ( x^(2/3) ) / (2 y) **

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Self-critique (if necessary):

Self-critique Rating:

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Question: Extra question. What is the simplified form of sqrt( 4 ( x+4)^2 ) and how did you get this result?

sqrt(4(x+4)(x+4))

=sqrt (2*2)(x+4)(x+4)

2 and x+4 can come outside the radicand

= 2 * (x+4)

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

We use two ideas in this solution:

sqrt(a b) = sqrt(a) * sqrt(b) and

sqrt(x^2) = | x |

To understand why sqrt(x^2) = | x | and not just x consider the following:

Let x = 5. Then sqrt(x^2) = sqrt( 5^2 ) = sqrt(25) = 5, so sqrt(x^2) = x.

It is also clear that in this case, | x | = | 5 | = 5, so | x | = x, and we can say that sqrt(x^2) = | x |.

Now let x = -5. We get sqrt(x^2) = sqrt( (-5)^2 ) = sqrt(25) = 5.

In this case sqrt(x^2) = 5 but x is not equal to 5, so sqrt(x^2) is not x.

However, in this case | x | = | -5 | = 5, so it is the case the sqrt(x^2) = | x |.

Using these ideas we get

sqrt( 4 ( x+4)^2 ) = sqrt(4) * sqrt( (x+4)^2 ) = 2 * | x+4 | **

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Self-critique (if necessary):

I see how you come up with the absolute value of x+4...Something I had never thought of..I just hope I can remember it

Self-critique Rating:

* Add comments on any surprises or insights you experienced as a result of this assignment.

Alot of surprises....This has been the hardest for me to grasp so far. Actually I dont feel I have completely grasped it,

Ill try to work on these more.

You appear to be doing well. I recommend more practice (the book gives you plenty of problems to practice on), and you are of course welcome to ask additional questions.