Assignment 8 R8

#$&*

course Mth 158

2/20/13 1:15PMVery sorry this is late, I thought I had submitted this assignment already. I didn't change anything from the way it was before.

008. `*   8

 

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

 

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Your solution:

 54^(1/3)

(27 * 2)^(1/3)

 (27)^(1/3) * 2^(1/3)

 3 * 2^(1/3)

 

 

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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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:

 (3 x y^2 / (81 x^4 y^2)^(1/3)

 (1 / 27 x^3)^(1/3)

 1 / 3x

 

 

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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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

 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 * ( 3 sqrt(6))

 6 sqrt(6^2) * 9 sqrt(6)

 36 * 9 sqrt(6)

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

 

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

 

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Your solution:

 R.8.36

 (sqrt(x) + sqrt(5) )^2

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

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

 x + 2 * sqrt (5x) + 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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Self-critique Rating: 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) / 2

 

 My textbook says R.8.42 is: 8xy - sqrt(25x^2y^2) + cube root of (8x^3y^3)

 8xy - sqrt(25x^2y^2) + cube root of (8x^3y^3)

 8xy - 5xy + 2xy

 5xy

 

 

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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

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

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

 sqrt(2) * (sqrt(7) - 2) / 3 

 

 

confidence rating #$&*: 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): OK

 

 

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Self-critique Rating: OK

 

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)^(1/2)

 

 

confidence rating #$&*: 1

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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):

 I knew the answer was x^(1/2), but I wasn’t sure how. I see now.

 

 

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Self-critique Rating: OK

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

 

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Your solution:

 25^(3/2)

 sqrt(25)^3

 5^3

 125

 

 

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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

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

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

 xy / x^(5/4) y^(2/4)

 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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: 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), defined for -3 < x < 3.

 

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Your Solution:

This is an incorrect solution, but this is what I did:

 ( (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) + x^2 / ( 9 - x^2) ^(3/2)

 

 

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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Self-critique (if necessary):

 I know what I did wrong and by reading through the given solution I managed to correct my problem and work it myself.

 

 

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Self-critique Rating: 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:

init vel = 0, height = 4

 v = sqrt (64 * 4 + 0^2)

 v = sqrt (256)

 v = 16

 init vel = 0, height = 16

 v = sqrt (64 * 16 + 0^2)

 v = sqrt(1024)

 v = 32

 init vel = 4, height = 2

 v = sqrt (64 * 2 + 4^2)

 v = sqrt (128 + 16)

 v = sqrt(144)

 v = 12

 

 

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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

 

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)

 cube root (24)

 8^(1/3) * 3^(1/3)

 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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

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

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

 5x^(2/3) / 2y

 

 

confidence rating #$&*: 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): OK

 

 

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Self-critique Rating: OK

 

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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)

 2x + 8

 

 

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):

 Is my answer correct even without the | | absolute value signs????

 

 

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Self-critique Rating: 1

@&

As long as x >= -4 your expression 2x + 8 would be non-negative and therefore would represent sqrt( 4 ( x + 4) ^ 2 ).

However if x < -4 this is not the case. For example if x = -5, you would have

sqrt(4 ( x + 4) ^ 2 ) = sqrt( 4 ( -5 + 4) ^ 2 ) = sqrt(4 * (-1)^2 ) = sqrt( 4 * 1) = sqrt(4) = 2,

whereas

2x + 8 = 2 * (-5) + 8 = -10 + 8 = -2.

Since 2x + 8 can be negative. whereas a square root cannot, your answer would need to be expressed as

| 2 x + 8 | .

*@

 

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

 

Math is tricky. Almost every time I think I know what I’m doing, I mess up, and every time I think what I’m doing is wrong, it’s actually right."

Self-critique (if necessary):

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Self-critique rating:

@&

There aren't that many rules, and for the most part they aren't hard to understand, but learning to use them correctly is always a challenge.

You're handling it very well. Keep up the good work.

*@