Assignment 15

course Mth 158

’¨åü”ó|•¹õ£}ÂÕŸz—«assignment #015

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

015. `query 15

College Algebra

07-02-2007

......!!!!!!!!...................................

15:20:23

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

......!!!!!!!!...................................

RESPONSE -->

First I subtracted 6 from both sides and got | 1 - 2z| = -6. Then I split the problem up into two equations and took away the absolute value signs. 1- 2z = 6 and 1- 2z = -6. I solved both equations and my solution set is { - 5/2 , 7 / 2}

confidence assessment: 2

.................................................

......!!!!!!!!...................................

15:23:00

** 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} **

......!!!!!!!!...................................

RESPONSE -->

The problem must have been a typo because it is equal to 0. I followed the same steps to get my answers in the other equation. I understand that there must be two seperate equations and how to solve them.

self critique assessment: 2

A typo is possible. If the right-hand side is indeed 0 then there is no solution, because an absolute value can't be negative.

.................................................

......!!!!!!!!...................................

15:31:39

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

......!!!!!!!!...................................

RESPONSE -->

I split it up into two problems.

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

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

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

Then I factored the equations.

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

Then I set each equation equal to 0.

x + 4 = 0 x - 1 = 0 x= 0 x+3 = 0

x = -4 x = 1 x =0 x = -3

My solution set is { -4 , 1, 0, -3 }

confidence assessment: 2

.................................................

......!!!!!!!!...................................

15:32:18

** 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. **

......!!!!!!!!...................................

RESPONSE -->

Those are the solutions I got.

self critique assessment: 3

.................................................

......!!!!!!!!...................................

15:37:10

1.6.36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.

......!!!!!!!!...................................

RESPONSE -->

I subtracted 3 from both sides | x+4| < 2

-2 < x + 4 < 2. Then subtracted 4, - 6 < x < -2. My solutions are -6 , -2

confidence assessment: 1

.................................................

......!!!!!!!!...................................

15:37:28

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

| x+4 | < 2

-2 < x+4 < 2

-6 < x < -2

......!!!!!!!!...................................

RESPONSE -->

That is exactly what I did.

self critique assessment: 3

.................................................

......!!!!!!!!...................................

15:40:58

1.6.48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.

......!!!!!!!!...................................

RESPONSE -->

- 1 >= -x -2 >= 1. Then I added 2, 1 >= -x >= 3.Since the x is negative switch the signs 1=< x =< 3.

confidence assessment: 2

.................................................

......!!!!!!!!...................................

15:43:00

**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}. **

......!!!!!!!!...................................

RESPONSE -->

I messed up on that problem. I forgot about infinty being a part of the solution. I understand now how to solve the problem and why the solution includes infinity.

self critique assessment: 2

................................................."

&#

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