course MTH 158
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13:18:01 **** query 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 --> first isolate the absolute value by subtracting 6 from both sides | 1 - 2z| = -6 1 - 2z = -6 or 1 -2z = 6 -2z = -7 or -2z = 5 z = 7/2 or z = -5/2
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13:18:58 ** 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 --> Right. The query problem =0 but this solution is to the problem in the book.
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13:22:20 **** query 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 subtract 2 f/both sides x^2 + 3x -4 = 0 or x^2 + 3x + 0 = 0 the second expression doesn't factor but the first does: (x+4)(x - 1)=0 x=-4,1 (-4, 1 )
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13:24:24 ** 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 --> yes, I forgot to factor out the second expression by the common monomial. I was looking at it as a trinomial which couldn't be factored. I see.
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13:26:12 **** query 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 --> first subtract 3 f/both sides | x + 4 | < 2 use rule for abs. value inequalities -2 < x + 4 < 2 -6 < x < -2 (-6, -2)
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13:26:19 STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2
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RESPONSE --> right
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13:27:59 **** query 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 --> -x - 2 <= -1 or -x - 2 >= 1 -x <= 1 or -x >= 3 x >= -1 or x <= -3
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13:28:50 **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 --> right
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