#$&* course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
.............................................
Given Solution: * * Starting with | 1-2z| +6 = 9 we add -6 to both sides to get | 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b: 1-2z=3 or 1-2z= -3 Solving both of these equations: -2z = 2 or -2z = -4 we get z= -1 or z = 2 We express our solution set as {-1, 2} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X^2 plus 3x minus 2 = +/-(2), Set up the + portion of the +/- solution x^2+3x-2=2...Add 2 to both sides then mover to the left and set = to zero to have x squared plus 3x minus 4=0..Factor the trinomial into (x+4)(x-1)=0 Set each the factors of the left hand side and set them to = zero and the set is 4,1 ..Now do the +/-...multiply -1 by the 2 inside the parentheees x squared plus 3 x - 2=-2 Factor out the GCF of x from each term in the polynomial x(x) plus x(3)=0 factor gcf x from x^2 plus 3x...x(xplus3)=0 Set to equal zero ..All sets: x=-4,1,0,-3 confidence rating #$&*: 3 OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: Alright since 3 does not have the variable to solve for, move to right and subtract 3..absolute x plus 4 is less than -3 plus 5.Add 5 to -3 to get 2 Remove the absolute value ….x=+/-x...x+4<+/-(2) ...X Plus 4 <2 ...X<-2...X+4>- (2)..Multiply -1 by the 2 inside the parentheses x+3>-2...Since 4 does not have the variable move to right by subtracting 4 from both sides...x>-4-2 Subtract 2 from -4 to get -6 x>-6...x<-2 and x>-6 ...-6
.............................................
Given Solution: * * STUDENT SOLUTION: | x+4| +3 < 5 | x+4 | < 2 -2 < x+4 < 2 -6 < x < -2 STUDENT QUESTION I was hoping to see more in the given solution as to why we move 2 to the left of the inequality. I think there is a formula for that, but I don’t remember what it is. Could you explain why we move the 2? INSTRUCTOR RESPONSE The 2 doesn't get moved. To understand what's going on: Think about the inequality | A | < = 4. This is clearly true if A = 4, 3, 2, 1 or 0. It's also clearly true if A = -1, -2, -3 or -4. It's not true if A = -5 or -6 or -7, etc.. So | A | < = 4 means the same thing as -4 <= A <= 4. More generally | A | < B says the same thing as - B < A < B. In your solution you said that | x + 4 | + 3 < 5 add -3 to both sides give us x + 4 < 2 This isn't so. The | | signs don't go away when you add -3 to both sides. You get | x + 4 | < 2, which means the same thing as -2 < x + 4 < 2 because of the rule we just say, that | A | < B means -B < A < B. Correcting your solution: | x + 4 | + 3 < 5 add -3 to both sides | x + 4 | < 2 add -2 to the left of the inequality -2 < x + 4 < 2 apply the rule for | A | < B with A = x + 4 and B = 2 -2-4 < x+4-4 < 2-4 simplify to get -6 < x < -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 OK ********************************************* 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 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. {-infinity, -3} U {-1, infinity}. confidence rating #$&*: 3 OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 3 OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!