#$&* course Math 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, because sometimes we get extraneous roots. But note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You factor and get: (v+6)(v+1) = 0. You will break it down into 2 little problems. Problem 1: v + 6 = 0. You will subtract 6 from both sides. Your answer is v = -6. Problem 2: v + 1 = 0. You will subtract 1 from both sides. Your answer is v = -1. Your answer is (-6, -1). You will plug it in to check your answer. (-6 + 6) (-1 + 1) = 0. You will get (0)(0) = 0. Your answer is correct. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: v^2+7v+6=0. This factors into (v + 1) (v + 6) = 0, which has solutions v + 1 = 0 and v + 6 = 0, giving us v = {-1, -6} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You will distribute to get x^2 + 4x = 12. You will subtract 12 from both sides. You will get x^2 +4x -12 = 0. You will factor to get (x+6)(x-2) = 0. You will set it up as two separate problems. Problem 1: x +6 = 0. You will subtract 6 from both sides, which gives you the answer x = -6. Problem 2: x - 2 = 0. You will add 2 to both sides. Your answer is x = 2. Your answer is (-6, 2). You will plug it in to check it: (-6 + 6)(2-2)=0. You get (0)(0) = 0. Your answer is correct: (-6, 2). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You will multiply both sides by the denominator x. You will get x^2 +12 = 7x. You will subtract 7x from both sides. You will get x^2 - 7x + 12 = 0. You will factor to get (x - 4)(x - 3)= 0. You will break it down into 2 problems. Problem 1: x - 4 = 0. You will add 4 to both sides. Your answer is x = 4. Problem 2: x - 3 = 0. You will add 3 to both sides. Your answer is x = 3. Your answer is (4, 3). You will plug it in to check. (4 -4)(3 - 3) = 0. You get (0)(0) = 0. You are correct. The answer is (4,3). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x + 12/x = 7 multiply both sides by the denominator x: x^2 + 12 = 7 x add -7x to both sides: x^2 -7x + 12 = 0 factor: (x - 3)(x - 4) = 0 apply the zero property x-3 = 0 or x-4 = 0 so that x = {3 , 4} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You will take the square root of 1. This gives you x+2 = + 1. You break it down into two problems. Problem 1: x +2 = 1. You will subtract 2 from both sides giving you x = -1. Problem 2: x + 2 = -1. You will subtract 2 from both sides and you get x = -3. Your answer is (-1, -3). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.38 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You must both sides by the denominator of 3. You will get 3x^2 + 2x -1 = 0. You will factor to get (3x - 1)(x + 1) = 0. You will break it down into 2 problems. Problem 1: 3x - 1 = 0. You will add 1 to both sides giving you 3x = 1. You will divide by 3 and get 1/3. Your answer is x = 1/3. Problem 2: x + 1 = 0. You will subtract 1 to both sides. Your answer is x = - 1. Your final answer is (1/3, -1) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * x^2 + 2/3x - 1/3 = 0. Multiply both sides by the common denominator 3 to get 3 x^2 + 2 x - 1 = 0. Factor to get (3x - 1) ( x + 1) = 0. Apply the zero property to get 3x - 1 = 0 or x + 1 = 0 so that x = 1/3 or x = -1. STUDENT QUESTION: The only thing that confuses me is the 1/3. Is that because of the 3x? INSTRUCTOR RESPONSE: You got the equation (3x - 1) ( x + 1) = 0. The product of two numbers can be zero only if one of the numbers is zero. So (3x - 1) ( x + 1) = 0 means that 3x - 1 = 0 or x + 1 = 0. You left out this step in your solution. x + 1 = 0 is an equation with solution x = -1 Thus the solution to our original equation is x = 1/3 or x = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.44 \ 52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You use the quadratic equation ax^2 +bx + c = 0. You will have a = 1, b = 6, c = 1. The quadratic formula is: x = -b + sqrt b^2 - 4ac/ 2a. You will plug in your numbers: x = -6 + sqrt 6^2 - 4 (1)(1) / 2(1). You will have -6 + (36 - 4)/ 2. You will have -6 + sqrt (32)/2. You have the solution of -6 + sqrt 32/2 and -6 - sqrt 32/2. Problem 1: -6 + sqrt of 16/2. This gives you -6 +sqrt 4/ 2. You will divide by 2. Your 1st answer is -3 +sqrt 2. Problem 2: -6 - sqrt 32/2. You will divide by 2. You have -3 - sqrt 16. This gives you -3 - sqrt 4. 36 - 4 = 32, so x has 2 real solutions, x = [-6 + sqrt(32) ] / 2 and x = [-6 - sqrt(32) ] / 2 Our solution set is therefore { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Now sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I am confused by how you have the (2). ------------------------------------------------ Self-critique Rating:
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Given Solution: * * Applying the quadratic formula with a = pi, b = 15 sqrt(2) and c = 20 we get x = [ (-15sqrt(2)) ± sqrt ( (-15sqrt(2))^2 -4(pi)(20) ) ] / ( 2 pi ). The discriminant (-15sqrt(2))^2 - 4(pi)(20) = 225 * 2 - 80 pi = 450 - 80 pi = 198.68 approx., so there are 2 unequal real solutions. Our expression is therefore x = [ (-15sqrt(2)) ± sqrt(198.68)] / ( 2 pi ). Evaluating with a calculator we get x = { -5.62, -1.13 }. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I need help where I got stuck. ------------------------------------------------ Self-critique Rating:
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Given Solution: * * Using x for the length of the shorter side of the rectangle the 2/1 ratio tells us that the length is 2x. If we cut a 1 ft square from each corner and fold the 'tabs' up to make a rectangular box we see that the 'tabs' are each 2 ft shorter than the sides of the sheet. So the sides of the box will have lengths x - 2 and 2x - 2, with measurements in feet. The box so formed will have height 1 so its volume will be volume = ht * width * length = 1(x - 2) ( 2x - 2). If the volume is to be 4 we get the equation 1(x - 2) ( 2x - 2) = 4. Applying the distributive law to the left-hand side we get 2x^2 - 6x + 4 = 4 Divided both sides by 2 we get x^2 - 3x +2 = 2. We solve by factoring. x^2 - 3x + 2 = (x - 2) ( x - 1) so we have (x - 2) (x - 1) = 2. Subtract 2 from both sides to get x^2 - 3 x = 0 the factor to get x(x-3) = 0. We conclude that x = 0 or x = 3. We check these results to see if they both make sense and find that x = 0 does not form a box, but x = 3 does. • So our solution to the equation is x = 3. x stands for the shorter side of the rectangle, which is therefore 3. The longer side is double the shorter, or 6. Thus to make the box: We take our 3 x 6 rectangle, cut out 1 ft corners and fold it up, giving us a box with dimensions 1 ft x 1 ft x 4 ft. This box has volume 4 cubic feet, confirming our solution to the problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1/2/100 / 1.2.108 \ 100 (was 1.3.96). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) s = -4.9 t^2 + 20 t; when 15 m high, when strike ground, when is ht 100 m. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Equation is -4.9t^2 + 20t = 15. Subtract 15 from each side. You have -4.9t^2 + 20t -15 = 0. You plug this into the quadratic formula. -20 + sqrt (20^2 - 4(-4.9) (-15)/ 2(-4.9) t = .99 and t = 3.09 You will plug your answers back in. confidence rating #$&*: 1 I am a little lost. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * To find the clock time at which the object is 15 m off the ground we set the height s equal to 15 to get the equation -4.9t^2 + 20t = 15 Subtracting 15 from both sides we get -4.9t^2 +20t - 15 = 0 so that t = { -20 ± sqrt [20^2 - 4(-4.9)(-15) ] } / 2(-4.9) Numerically these simplify to t = .99 and t = 3.09. Interpretation: • The object passes the 15 m height on the way up at t = 99 and again on the way down at t = 3.09. To find when the object strikes the ground we set s = 0 to get the equation -4.9t^2 + 20t = 0 which we solve to get t = [ -20 ± sqrt [20^2 - 4(-4.9)(0)] ] / 2(-4.9) This simplifies to t = [ -20 +-sqrt(20^2) ] / (2 * -4.9) = [-20 +- 20 ] / (-9.8). The solutions simplify to t = 0 and t = 4.1 approx. Interpretation: The object is at the ground at t = 0, when it starts up, and at t = 4.1 seconds, when it again strikes the ground. To find when the altitude is 100 we set s = 100 to get -4.9t^2 + 20t = 100. Subtracting 100 from both sides we obtain -4.9t^2 +20t - 100 = 0 which we solve using the quadratic formula. We get t = [ -20 ± sqrt (20^2 - 4(-4.9)(-100)) ] / 2(-4.9) The discriminant is 20^2 - 4 * -4.9 * -100, which turns out to be negative so we do not obtain a solution. Interpretation: We conclude that this object will not rise 100 ft. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: