Assignment 8

course mth 158

6/29

008. `* 8

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

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Your solution:

cube root of 54 = cube root of 27 *2

Cube root of 54 = cube root of 3^3 *2

Cube root of 54 = 3 * cube root of 2

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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):

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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:

[(3xy^2) / (81 x^4 y^2)]^1/3

[(3)/(81x^2)]^1/3

[(1 / (27x^2)]^1/3

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

1 / 3*x

1/3x

confidence rating #$&*2

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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):

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

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Your solution:

2 sqrt (12) - 3 sqrt (27)

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

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

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

4 sqrt (3) - 9 sqrt (3)

- 5 sqrt (3)

confidence rating #$&*3

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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:

(2 sqrt(6) + 3) (3 sqrt(6))

(2 sqrt(6) * 3 sqrt (6) + 3*3sqrt(6))

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

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

6 sqrt (36) + 9 sqrt (6)

6 * 6 +9 sqrt (6)

36 + 9 sqrt (6)

confidence rating #$&*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):

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

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Your solution:

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

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

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

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

confidence rating #$&*3

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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):OK

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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)

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

3 sqrt (2) / (sqrt 2)^2

(3 sqrt(2)) / 2

You must multiply by the denominator. Since sqrt (2) / sqrt (2) = 1, it does not change the answer.

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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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Self-critique (if necessary):

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

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

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

sqrt (2) * (sqrt(7) - 2) / 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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Self-critique (if necessary):

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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)^(1/6)

x^(3*1/6)

x^1/2

The only way to simplify would be to multiply the exponents, since you cannot find a 6th root of x^3

easily. 3 * 1/6 = 1/2.

confidence rating #$&*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):

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

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Your solution:

25 = 5^2

(5^2)3/2

5^(2 * 3/2)

5^(6/2)

5^3

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Given Solution:

25^(3/2) =

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

5^(2 * 3/2) =

5^3.

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Self-critique (if necessary):

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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:

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

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

x^5/4 * y^5/4 / x^6/4 y^3/4

(x^5/4-6/4) * y^5/4-3/4

x^-1/4 * y^2/4

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

confidence rating #$&*2

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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 + Ľ? 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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Self-critique (if necessary):OK

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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:

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

(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

confidence rating #$&*1

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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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Self-critique (if necessary):

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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:

v1:

sqrt (64*4 + 0^2)

sqrt (256)

sqrt (256) = 16

v2:

sqrt (64 * 16 + 0^2)

sqrt (1028) = 32

v3:

sqrt (64 * 2 + 4 ft/s)

sqrt (16 * 4 * 2 + 2^2 ft/s)

4 * 2* 2ft/s sqrt (2)

16 sqrt (2) ft/s

confidence rating #$&*2

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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):

the last one threw me. I thought the units attached to the +4 would make it impossible to group with the

others until after you had removed it from the radical.

64 actually has units as well. The units would be ft / s^2.

(4 ft / s)^2 would be 16 ft^2 / s^2, which would agree with the units of 64 ft/s^2 * 2 ft.

So you would have like terms in the radical.

The way the formula is stated you don't actually need to carry the units through the calculation, but it's an excellent practice to do so. I'm glad you're thinking about the units.

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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)

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

2 * 3^(1/3)

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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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Self-critique (if necessary):

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

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

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

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

5(x^2/3) / 2y

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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Self-critique (if necessary):

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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:

sqrt( 4 ( 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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Self-critique (if necessary):

I did not consider the possiblity of x being a negative number.

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

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

#$&*