Assignment 14

course MTH 158

D|HƒۥÁassignment #014

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014. `query 14

College Algebra

06-24-2008

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10:26:33

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 -->

My book has the question as |1-2z| + 6 = 9, which makes sense because then subtracting the 6 from both sides would make the inequality equal to 3. However, without that |1-2z| + 6 being equal to nine, after subtracting the 6 from both sides, that makes |1-2z| = -6, and it's impossible to have a negative absolute value, so I am just completely confused. However, I'll try to solve it...

|1-2z| = -6 will mean that 1 -2z = -6 or 1 -2z = 6

-2z = -6 or -2z = 5

-2z / -2 = -6 / -2 -2z / -2 = 5 / -2

z = 3 or z = - 5/2

{- 5/2, 3}

confidence assessment: 1

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10:28:03

** 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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RESPONSE -->

So it was a typo... I thought so. Will I still have the consequence?

self critique assessment: 1

you used the right idea in solving the equation (the incorrect equation, but that was my fault). However note one thing related to your solution, which does not occur in the given solution.

| a | = b has no solution if b < 0, since an absolute value can't be negative.

So |1 - 2 z | = -6 has no solution.

Had the equation been | 1 - 2 z | = 6, then everything you said would be so and the equation would have had one of the two solutions you specified.

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10:40:12

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

x = {-4, 1} or x = {-3, 0}

{-4, -3, 1, 0}

confidence assessment: 2

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10:40:29

** 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: 2

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10:43:54

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 -->

| x + 4 | + 3 < 5

|x + 4| < 2 so x + 4 < 2 or x + 4 < -2

x < -2 or x < -6

confidence assessment: 2

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10:47:07

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

| x+4 | < 2

-2 < x+4 < 2

-6 < x < -2

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RESPONSE -->

I see.

self critique assessment: 2

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10:53:12

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 -->

Using the rule |u| >= a is equivalent to u <= -a or u >= a

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

-x <= 1 or -x >= 3

x >= -1 or x <= -3

{x | x >= -1 or x <= -3}

confidence assessment: 2

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10:53:36

**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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RESPONSE -->

ok

self critique assessment: 3

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You're doing really well.

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Let me know if you have questions. &#