Assignment 14

course MTH 158

ܦ’`èÓÈó[à¹ZˆÙ©¦`’ïÙ±ïþСîz¡¦~C›assignment #014

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.

014. `query 14

College Algebra

10-09-2007

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16:57:13

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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RESPONSE -->

1-2z+6=9 or 1-2z+6=-9

-2z=2 -2z=-16

z=-1 z=8

{-1,8}

confidence assessment: 3

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16:57:23

10-09-2007 16:57:23

** 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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NOTES ------->

Be sure to isolate the absolute value on one side before you split into two equations.

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17:03:49

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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RESPONSE -->

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

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

x^2+4x-1x-4=0 x(x+3)=0

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

x=1 x=-4

{1,-4,0,-3}

confidence assessment: 3

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17:04:01

** 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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RESPONSE -->

ok

self critique assessment: 3

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17:05:35

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

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RESPONSE -->

-5

You can't combine the 4 in the abs value with the 3 outside. Again, isolate the abs value expression before splitting into two inequalities.

-5

-12

-12

confidence assessment: 3

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17:05:48

10-09-2007 17:05:48

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

| x+4 | < 2

-2 < x+4 < 2

-6 < x < -2

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NOTES ------->

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17:09:49

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

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RESPONSE -->

-1<=-x-2>=1

-1<=-x-2 -x-2>=1

1<=-x -x>=3

-1>=x x<=-3

{x/x<=-3>=-1}

in the interval, of ( ,-3) and (-1, )

confidence assessment: 2

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17:09:59

10-09-2007 17:09:59

**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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NOTES ------->

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See my notes and let me know if you have questions.