Assignment 14

course MTH 158

7/4/09 1:20 A.M.

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.

014. `* 14

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Self-critique (if necessary):

Self-critique Rating:

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Question: * 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.

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Your solution:

setting up 2 different equations we can have

1-2z = -6

-2z=-7

z=-7/-2

z=7/2

second equation

1-2z=6

-2z = 5

-2z/-2 = 5/-2

z = -5/2

(-5/2,7/2)

Confidence rating: 3

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Given Solution:

* * Starting with

| 1-2z| +6 = 9 we add -6 to both sides to get

| 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b:

1-2z=3 or 1-2z= -3 Solving both of these equations:

-2z = 2 or -2z = -4 we get

z= -1 or z = 2 We express our solution set as

{-2/3,2} **

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Self-critique (if necessary):

I worked the equation the way you had it wrote in the question

|1 - 2z| + 6 = 0

so wouldnt our 2 equations be

1 - 2z = -6 and 1 - 2z = 6

That would be correct for the equation in the problem as stated above.

However the right-hand side should have been 9, and the given solution is correct for the correct equation.

You clearly understand it either way.

Self-critique Rating:

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Question: * 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2

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Your solution:

two equations

x^2 + 3x - 2 = 2

x^2 + 3x - 4 + 0

(x+4)(x+1)

x = -4

x = 1

x^2 + 3x -2 = -2

x^2 + 3x -2 + 2 = 0

x^2 + 3x = 0

x( x+3) = 0

x= 0

x = -3

Confidence rating: 3

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Given Solution:

* * My note here might be incorrect.

If the equation is | x^2 +3x -2 | = 2 then we have

x^2 + 3x - 2 = 2 or x^2 + 3x - 2 = -2.

In the first case we get x^2 + 3x - 4 = 0, which factors into (x-1)(x+4) = 0 with solutions x = 1 and x = -4.

In the second case we have x^2 + 3x = 0, which factors into x(x+3) = 0, with solutions x = 0 and x = -3. **

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Self-critique (if necessary):

Self-critique Rating:

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Question: * 1.6.40 \ 36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.

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Your solution:

2 equations

| x + 4 | + 3 < 5

| x + 4 | < 5 - 3

| x + 4 | < 2

so we have x+4 < 2 and x + 4 < -2

which equals x < -2 and x < -6

(-6 < x < -2)

Confidence rating: 3

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Given Solution:

* * STUDENT SOLUTION: | x+4| +3 < 5

| x+4 | < 2

-2 < x+4 < 2

-6 < x < -2

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Self-critique (if necessary):

Self-critique Rating:

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Question: * 1.6.52 \ 48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.

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Your solution:

| - x - 2 | >= 1 or |-x - 2| >= -1

Dividing -1 from both sides -x >= 3 (x =< -3)

In the second equation diving -1 from both sides -x >= 1 (x=< -1)

Confidence rating: 2

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Given Solution:

* * Correct solution:

| -x -2 | >= 1 Since | a | > b means a > b or a < -b (note the word 'or') we have

-x-2 >= 1 or -x -2 <= -1. These inequalities are easily solved to get

-x >= 3 or -x <= 1 or

x <= -3 or x >= -1.

So our solution is

{-infinity, -3} U {-1, infinity}. **

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Self-critique (if necessary):

Self-critique Rating:

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