Assigment 14 16

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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 = 9.

 

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Your solution:

 | 1 - 2 z | + 6 = 9

1 - 2 z = 3

-2z = 2

z = -1

 or

 1 - 2 z = -3

 -2z = -4

 z = 2

 {-1, 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

 

{-1, 2} **

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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:

 | x^2 +3x - 2 | = 2

 x^2 +3x - 2 = 2

 x^2 + 3x - 4 = 0

 (x + 4)(x - 1) = 0

 {-4 , 1}

 or

 | x^2 +3x - 2 | = -2

 x^2 +3x - 2 = -2

 x^2 + 3x = 0

 x(x + 3) = 0

 {0 , -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): OK

 

 

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Self-critique Rating: OK

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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:

 | x + 4 | + 3 < 5

 x + 4 < 2

 -2 < x + 4 < 2

 -6 < x < -2

 (-6 , -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

 

STUDENT QUESTION

 

I was hoping to see more in the given solution as to why we move 2 to the left of the inequality. I think there is a formula for that, but I don’t remember what it is.

Could you explain why we move the 2?
INSTRUCTOR RESPONSE

 

The 2 doesn't get moved.  To understand what's going on:
Think about the inequality 
| A | < = 4.
This is clearly true if A = 4, 3, 2, 1 or 0.
It's also clearly true if A = -1, -2, -3 or -4.
It's not true if A = -5 or -6 or -7, etc..
So
| A | < = 4 means the same thing as-4 <= A <= 4.
More generally
| A | < B says the same thing as
- B < A < B.
In your solution you said that
| x + 4 | + 3 < 5 add -3 to both sides give usx + 4 < 2 
This isn't so. The | | signs don't go away when you add -3 to both sides. You get
| x + 4 | < 2, which means the same thing as-2 < x + 4 < 2 because of the rule we just say, that | A | < B means -B < A < B.
Correcting your solution:
| x + 4 | + 3 < 5 add -3 to both sides| x + 4 | < 2 add -2 to the left of the inequality -2 < x + 4 < 2 apply the rule for | A | < B with A = x + 4 and B = 2-2-4 < x+4-4 < 2-4 simplify to get-6 < x < -2

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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

 -x - 2 >= 1

 -x >= 3

 x <= -3

 or

 | -x - 2 | <= -1

 -x - 2 <= -1

 -x <= 1

 x >= -1

 (-∞ , -3] U [-1 , ∞)

 

confidence rating #$&*: 3

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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):

 Is the interval notation in my solution correct?

@&

Assuming that ∞ is the symbol for infinity, your interval solution is correct.

You would type this simply as

(-infinity, -3) U (-1, infinity).

*@

 

 

 

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#