assignment 14

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.

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

17:20:10 **** query 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.

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

RESPONSE --> | 1 - 2z | + 6 = 0 1 + 2z + 6 = 0 7 + 2z = 0 2z = -7 z = -7/2

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

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

17:20:44 ** 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 --> opps I thought I had to solve for x for some reason. I know what I did wrong.

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

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

20:32:18 **** query 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2

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

RESPONSE --> | x^2 +3x - 2 | = 2 x^2 + 3x - 2 = 2 x^2 + 3x - 4 = 0 x^2 + 3x = -4 x(x+3)=-4 x = -4 x = -7 or x^2 + 3x - 2 = -2 x^2 + 3 = 0 x^2 = -3 x = +-sqrt(3)

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

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

20:34:11 ** 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 --> I tried to factor this one, but for some reason I couldn't get it. After reading your note, I figured it out.

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

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

20:37:36 **** query 1.6.36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.

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

RESPONSE --> |x + 4| + 3 < 5 x + 4 + 3 < -5 x + 7 < -5 x < -12 or x + 4 + 3 >5 x + 7 > 5 x > -2

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

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

20:38:53 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2

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

RESPONSE --> My mistake was that I didn't subtract 3 from both sides, indstead I left it and added it to 4.

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

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

20:45:05 **** query 1.6.48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.

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

RESPONSE --> | -x -2 | >= 1 -x - 2 >= 1 -x >= 3 x >= -3 or -x - 2 <= -1 -x <= 1 x <= -1

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

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

20:46:52 **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 didn't do the other side, other than that I got it correct.

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

"

Your work on this assignment is very good. Let me know if you have questions.