assignment 14

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course Mth 158

10/6 5

014. `*   14*********************************************

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 isolate the absolute value be subtracting 6 from both sides

| 1 - 2z| = 3 now you get 1 - 2z = 3 or 1 - 2z = -3 subtract one from both sides on both equations

- 2z = 2 or - 2z = -4 divide both sides of both equations by -2 and get

z=-1 or z=2 aka {-1,2}

confidence rating #$&*: ok

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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 you get

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

x^2 +3x -4= 0 or x^2 +3x = 0

(x-1)(x+4)=0 or x(x+3)=0

x=1, -4 or x=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 subtract 3 from both sides

| x + 4 | < 2

-2

-6

{x|-6

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 us

x + 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 or -x - 2 <= -1 add 2 to both sides on both equations

-x>=3 or -x<=1 divide both sides of both equations by -1

x<=-3 or x>=-1

{x|x<=-3 or x>=-1} aka (-infinity, -3] or [-1, infinity)

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):ok

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Self-critique Rating:ok

&#This looks very good. Let me know if you have any questions. &#