course mth158 10/1/09 @11:55 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: * * 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} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot the rule that when you have a | = b means that a = b or a = -b: and I didn’t solve for the negative. ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 + 3x – 2 - 2= 0 x^2 + 3x – 2 – (-2) = 0 x^2 + 3x – 4 =0 x^2 + 3x = 0 (x + 4) (x – 1) = 0 x(x+3) = 0 x = -4 x=1 x= -3 x= 0 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.6.40 \ 36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | x + 4 | < 2 -2 < x + 4< 2 -2 – 4 < x < 2 – 4 -6 < x < -2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.6.52 \ 48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -x – 2 > 1 -x – 2 < -1 -x > 3 -x< 1 x < -3 x> -1 (-infinity, -3] [-1, infinity) ---------] [----------- -3 -1 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: "