Assignment 8

#$&*

course Math 158

7:27 p.m. 6/4/12

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: I would a factor tree. I would use 9 and 6. 9 breaks down into 3*3 and 6 down in 3*2. Your cube is 3. Answer is 3 cube root sign and 2 inside.

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

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Your solution: I would have factored it down to 3xy^2/ 3*3*3*3x^4 y^2. Final answer: 1/3x are left after cancelling and making cubes.

confidence rating #$&*: 3

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

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Your solution: 2sqrt (2*2*3) -3sqrt(3*3*3) gives you 2sqrt(2^2*3) -3(3^3) (2*2-3*3)/3 Final answer: -5 sqrt 3

confidence rating #$&*: )

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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)

= -5 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: Factor out. 36+ 9sqrt 6. Section off and multiply after you factor.

confidence rating #$&*: 2 1/2

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

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Your solution: (sqrt(x) + sqrt(5) * (sqrt(x) + sqrt(5)

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

Sqrt(x)*sqrt(x) + sqrt(5) *sqrt(5) *I cannot see where the extra x and 5 comes from. Can you explain it from here?

@&

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

is not

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

for the same reason that

(a + b) * (a + b) is not a^2 + b^2.

(a + b) * ( a + b) = a * ( a + b) + b * (a + b) = a^2 + ab + ba + b^2.

Since ab = ba, this is a^2 + 2 a b + b^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

as indicated in the given solution.

*@

confidence rating #$&*: I am lost at this point.

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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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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: 3*sqrt 2 *sqrt2/sqrt2 = 3sqrt2/2 We are multiplying by the same numerator and denominator.

confidence rating #$&*: 3

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Given Solution:

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

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

(3 sqrt(2) ) /2.

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

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Your solution: sqrt 2/(sqrt 7 +2) *(sqrt7 -2)/(sqrt7-2) = 7 -2sqrt7 +2sqrt7 -4 = sqrt2(sqrt7-2)/7-4 = sqrt14-2sqrt2/3 You distribute to get the final answer.

confidence rating #$&*: 3

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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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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: x*3^/6 reduces ½. x^1/2 This gives you power/root.

confidence rating #$&*: 3

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

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Your solution: 25^3/2 = sqrt 25= sqrt 25*25*25 = 25sqrt25= the sqrt 25 = 5*5 = 25*5 = 125= 5^3

confidence rating #$&*: 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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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).

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Your solution: You have to multiply the exponents to get: x^1/4 y^1/4 x y/ x^3/2 y^3/4 = x^(1/4+1) y(1/4+1)/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^-1/4 y^1/2 = y^1/2/x^1/4

confidence rating #$&*: 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).

STUDENT QUESTION

I wrote the entire given solution on paper to see how to solve, but I am still confused when it gets to the

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

How do you get 1 + ¼? Does the 1 come from the xy on the right of the numerator?

INSTRUCTOR RESPONSE

The numerator of the expression

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

contains two factors which are powers of x. The two are

x^(1/4) and x^1 (the latter could be written just as x, but to apply the laws of exponents it's not a bad idea to write the exponent explicitly).

When you multiply these two factors, the laws of exponent tell you that you get x^(1/4 + 1) = x^(5/4).

The same thing happens with the y factors.

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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), defined for -3 < x < 3.

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Your Solution: (9-x^2)^1/2 + x^2(9-x^2)^-1/2)/9-x^2 = (9-x^2)^1/2 + x^2/(9-x^2)^1/2/9-x^2

confidence rating #$&*: 2

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Given Solution:

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

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

[ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ] =

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

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

In the third step the exponent ^1 on the (9 - x^2) expressions wasn't necessary, but was included to explicitly show the exponents and the application of the laws of exponents.

The first term in the 4th step is obtained as follows:

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

EXPANDED EXPLANATION OF STEPS

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

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

In the above step we have replace (9 - x^2) ^ (-1/2) in the numerator by (9 - x^2)^(1/2) in the denominator, following the rule that a^-b = 1 / (a^b) with a = (9 - x^2) and b = 1/2.

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

[ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ]

The above step is just the distributive law of multiplication over addition, in which we multiply through the expression ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) by 1 / (9 - x^2). The brackets [ ] have been added to clarify the two terms in the resulting expression, but the expression has the same meaning without them.

[ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ] =

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

(9 - x^2) ^ (1/2) / (9 - x^2)^2 = (9 - x^2)^-1/2, by the laws of exponents; and (9 - x^2)^(1/2) * (9 - x^2) = (9 - x^2) ^(3/2) by the laws of exponents.

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

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

(9 - x^2) ^(-1/2) has been replaced by 1 / (9 - x^2) ^(1/2), using a^-b = 1 / a^b.

All the exponents in the final expression are positive.

It would also be possible to factor out 1 / (9 - x^2)^(1/2), though this wasn't requested and isn't necessary in the problem as stated. The result would be

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

This could be further simplified to

1 / (9 - x^2)^(1/2) * ( 9 / (9 - x^2) ) , which is equal to

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

You aren't expected to be able to read these expressions. You are expected to be able to write them out in standard form; having done so you should understand.

However these expressions are fairly challenging, so some of the expressions will be depicted here

( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) would be depicted in standard notation as

(9 - x^2) ^(-1/2) + x^2 / (9 - x^2)^(3/2) would be depicted in standard notation as

1 / (9 - x^2)^(1/2) + x^2 / (9 - x^2)^(3/2) would be depicted in standard notation as

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@&

The expression

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

isn't quite grouped correctly (that last 9 - x^2 should be grouped). The exponent 1/2 also needs to be grouped. In addition, since you have a complex fraction for the last term, you might want to keep things clear by grouping it as

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

The correct grouping would be

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

*@

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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: v= sqrt64h + 0^2

V=0, h=4 = sqrt64(4) +0 = sqrt 256 = 16

V=0, h=16 = sqrt 64(16) + 0 = sqrt 1024 = 32

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

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Your solution: (24)^1/3 = 3sqrt24. Break 24 down to 2*2*2*3 = 2^3sqrt3 = 2* 3^1/3

confidence rating #$&*: 3

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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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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 y)^1/3 (125x^3)^1/3/(8x^3y^4)^1/3 = x^2/3 y^1/3 (125)^1/3 x/ 8^1/3x^4/3 = x^(2/3+1-1) y^(1/3-4/3) *5/2 = 5x^2/3/2y

@&

Correct grouping would be

5x^(2/3) / 2y

Without the grouping x would be raised to the power 2, not the power 2/3.

*@

confidence rating #$&*: 3

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

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Your solution: 4(x+)^2= 4(x+4)(x+4) = (4x+16)(x+4)= 4x(x+4) + 16(x+4) = 4x^2 +16x+16x + 64 = 4x^2 + 32x +64

@&

Good, but there is also the sqrt to be considered.

sqrt(4x^2 + 32x +64) = sqrt(4 * (x^2 + 8 x + 16) ) = sqrt(4 * (x + 4)^2) = sqrt(4) * sqrt( (x + 4)^2 ) = 2 * sqrt(x + 4)^2 = 2 | x + 4 | .

*@

confidence rating #$&*: 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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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#