#$&* course Mth 158 2/02 130I am submitting Assignment 1.1 and 1.2 in this one submission 1.1
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation in the above note is 5y + 6 = -18 - y. The solution to this equation is found by practically the same steps but you end up with y = -4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2x+1) / 3 + 16 = 3x multiply both sides by 3 2x +49 = 9x subtract 2x from both sides 49 = 7x now divide both sides by 7 x = 7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides to get 49 = 7x Divide both sides by 7 to get x = 7. STUDENT QUESTION I was wondering at the end since it ended up 49 = 7x and you divide by 7 and say x = 7 would you have to make it a -7 if you move it to the opposite side of the equation? INSTRUCTOR RESPONSE It's not a matter of 'moving things around', but a matter of adding or subtracting the same quantity on both sides, or multiplying or dividing both sides by the same quantity. In this case both sides are divided by 7, which doesn't involve any negative signs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+2)(x-3) = (x+3)^2 use the distributive law to expand x^2 - x 6 = x^2 + 6x + 9 subtract x^2 from both sides -x 6 = 6x + 9 add x to both sides -6 = 7x + 9 subtract 9 from both sides -15 = 7x divide both sides by 7 x = -15/7 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 x = -15/7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/ YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9) first we factor (x^2 9) x / (x+3) (x-3) + 4 / (x+3) = 3 / (x+3) (x-3) now we have to find the (LCD) x * (x+3) (x-3) / (x+3) (x-3) + 4 * (x+3) (x-3) = 3 * (x+3) (x-3) / (x+3) (x-3) now we cancel out x + 4*(x-3) = 3 use the distributive law x + 4x 12 = 3 now add 12 to both sides and combine like terms 5x = 15 divide 5 x = 3 If 3 is substituted into the original the original equation the answer we get is undefined because you cannot divide by 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Apply the Distributive Law, rearrange and solve: x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. STUDENT COMMENT x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) Since you have like terms (x^2 9) on both sides, they cancel each other out INSTRUCTOR RESPONSE If something 'cancels' by multiplication or division, it has to 'cancel' from all terms. (x^2 - 9) is not a multiplicative or divisive factor of the term 4 / (x + 3) so that factor does not 'cancel'. You can multiply or divide both sides by the same quantity, or add and subtract the same quantity from both sides. Anything called 'cancellation' that doesn't result from these operations is invalid. Because 'cancellation' errors are so common among students at this level, my solutions never mention anything called 'cancellation'. If you multiply both sides of the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) by (x^2 - 9), you get ( x / (x^2-9) + 4 / (x+3) ) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9) so that x / (x^2-9) * (x^2 - 9) + 4 / (x+3) * (x^2 - 9) = 3 / (x^2-9) * (x^2 - 9). The (x^2 - 9) does then 'cancel' from two of the three terms, but not from the third. You get x + 4 / (x+3) * (x^2 - 9) = 3. You're still stuck with an x^2 - 9 factor on one of the terms, and a denominator x - 3. However this equation does represent progress. If you factor x^2 - 9 into (x-3)(x+3), things quickly simplify. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7) first we need to find our common denominator (LCD) = (5w + 7) (10w-7) now we multiply both sides by the (LCD) (8w + 5) * (5w + 7) * (10w-7) / (10w - 7) = (5w + 7) * (10w-7) * (4w - 3) / (5w + 7) now we can cancel out like terms (8w + 5) * (5w + 7) = (10w-7) * (4w - 3) now we use the distributive law to combine the terms 40w^2 + 81w + 35 = 40w^2 58w + 21 now we will subtract 40w^2 from both sides 81w + 35 = -58w + 21 now we will add 58w to both sides 139w + 35 = 21 now subtract 35 from both sides 139w = -14 now divide by 139 w = -14/139 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * GOOD STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) STUDENT QUESTION: (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..; INSTRUCTOR RESPONSE: It doesn't look like you used the distributive law to multiply those binomials. (5w+7)(8w+5) = 5w ( 8w + 5) + 7 ( 8w + 5)= 40 w^2 + 25 w + 56 w + 35 = 40 w^2 + 81 w + 35. (10w-7)(4w-3) = 10 w ( 4 w - 3) - 7 ( 4 w - 3) = 40 w^2 - 30 w - 28 w + 21 = 40 w^2 - 58 w + 21. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 - a x = b, a <> 0 first we will subtract 1 from both sides -ax = b 1 mp we divide both sides by a x = (b 1) / -a multiply the right side by -1 x = (1 b) / a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * extra problem (was 1.1.72). Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^3 + 6 x^2 - 7 x = 0 first factor out the x x (x^2 + 6x 7) = 0 now factor the trinomial x (x + 7) (x 1) now set them = to 0 to solve x = 0 or x +7 = 0 , x = -7 or x 1 = 0 , x = 1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we need to find the average of our test grades 86 + 80 + 84 + 90 = 340 now we divide by 4 340 / 4 = 85 this the first part of his grade since in the problem it said that the exam is worth 2/3s of his grade which x will represent (2 * x + 85) / 3 now we have to see what we need to make a B (2 * x + 85) / 3 = 80 first we multiply both sides by 3 2 * x + 85 = 240 now we subtract by 85 2x = 155 now we divide by 2 x = 77.5 now we have to find out what we need to make and A (2 * x + 85) / 3 = 90 multiply by 3 (2 * x + 85) = 270 subtract by 85 2x = 185 divide by 2 x = 92.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = -g t + v0 for t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = -g t + v0 solve for t v = -g t + v0 first we subtract v0 v v0 = -gt now we divide by g (v v0) / -g = t now multiply the left side by -1 t = (v0 v) / g confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = -g t + v0, add -v0 to both sides to get v - v0 = -gt. Divide both sides by -g to get (v - v0) / (-g) = t so that t = -(v - v0) / g = (-v + v0) / g. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok * Add comments on any surprises or insights you experienced as a result of this assignment. I found this assignment easy and pretty self explanatory Assignment 1.2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 011. `* 11 * 1.2.13 \ 5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z^2 - z - 6 = 0 first we need to factor (z 3) (z + 2) = 0 now we set them equal to 0 z 3 = 0 or z + 2 = 0 when we solve for z we get z = {3, -2} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v^2+7v+6=0 first we factor (v + 6) (v + 1) now we set them to 0 v + 6 = 0 and v + 1 = 0 now we solve for v v = {-6, -1} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: v^2+7v+6=0. This factors into (v + 1) (v + 6) = 0, which has solutions v + 1 = 0 or v + 6 = 0, giving us v = {-1, -6} &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x(x+4)=12 first we need to turn this equation into a trinomial = 0 x^2 + 4x 12 = 0 now we have to factor (x + 6) (x - 2) = 0 now set each to 0 x + 6 = 0 or x 2 = 0 no we solve for x x = {-6, 2} confidence rating #$&*: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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x + 12/x = 7 first we get rid of the x from the denominator by multiplying both sides by x x^2 + 12 = 7x now we need to make it a trinomial = 0 x^2 7x + 12 = 0 now we factor (x 3)(x 4) = 0 now we set both to 0 x 3 = 0 or x 4 = 0 now solve for x x = {3, 4} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x + 12/x = 7 multiply both sides by the denominator x: x^2 + 12 = 7 x add -7x to both sides: x^2 -7x + 12 = 0 factor: (x - 3)(x - 4) = 0 apply the zero property x - 3 = 0 or x - 4 = 0 so that x = {3, 4} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+2)^2 = 1 first we need to find the sqrt of both sides x + 2 = +-sqrt (1) x = -2 1 or x = -2 + 1 now solve for x x = {-3, -1} confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 1.2.44 (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: x^2 + 2/3 x - 1/3 = 0 first we multiply by the denominator of 3 3x^2 + 2x 1 = 0 now we factor (3x 1) (x + 1) = 0 now we set both to 0 3x 1 = 0 or x + 1 = 0 solve for x x = 1/3 or x = -1 x = {1/3, -1} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * x^2 + 2/3x - 1/3 = 0. Multiply both sides by the common denominator 3 to get 3 x^2 + 2 x - 1 = 0. Factor to get (3x - 1) ( x + 1) = 0. Apply the zero property to get 3x - 1 = 0 or x + 1 = 0 so that x = 1/3 or x = -1. STUDENT QUESTION: The only thing that confuses me is the 1/3. Is that because of the 3x? INSTRUCTOR RESPONSE: You got the equation (3x - 1) ( x + 1) = 0. The product of two numbers can be zero only if one of the numbers is zero. So (3x - 1) ( x + 1) = 0 means that 3x - 1 = 0 or x + 1 = 0. You left out this step in your solution. x + 1 = 0 is an equation with solution x = -1 Thus the solution to our original equation is x = 1/3 or x = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I doubted myself ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.2.50 \ 52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x ^2 + 6x + 1 = 0 first we need to find the values to be substituted into the quadratic formula x = [-b +- sqrt (b^2 4ac)] / 2a a=1, b=6, c=1 now we need to substitute into the quadratic formula [-6 +- sqrt (6^2 4 * 1 * 1)] / 2 * 1 = [-6 +- sqrt (36 - 4)]/ 2 = [-6 +- sqrt (32)] / 2 = this can be simplified [-6 +- sqrt (2 * 16)] / 2 = [-6 +- 4 sqrt (2)] / 2 = [-3 +- 2 sqrt (2)] = {[-3 + 2 sqrt (2)], [-3 2 sqrt (2)]} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^2 + 6x + 1 = 0 we identify our equation as a quadratic equation, having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ± sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ± sqrt(36 - 4) / 2 x = { -6 ± sqrt (32) ] / 2 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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.2.78 \ 72 (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: pi x^2 + 15 sqrt(2) x + 20 = 0 we first need to find the values to be substituted a = pi, b = 15 sqrt (2), c = 20 x = [-b +- sqrt (b^2 4ac)] / 2a x = [ (-15sqrt(2)) +- sqrt ( (-15sqrt(2))^2 -4(pi)(20) ) ] / ( 2 pi ) = approximate value x = [ (-15sqrt(2)) +- sqrt (198.68)] / 2 pi = to unequal answers x = {-5.62, -1.13} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Applying the quadratic formula with a = pi, b = 15 sqrt(2) and c = 20 we get x = [ (-15sqrt(2)) ± sqrt ( (-15sqrt(2))^2 -4(pi)(20) ) ] / ( 2 pi ). The discriminant (-15sqrt(2))^2 - 4(pi)(20) = 225 * 2 - 80 pi = 450 - 80 pi = 198.68 approx., so there are 2 unequal real solutions. Our expression is therefore x = [ (-15sqrt(2)) ± sqrt(198.68)] / ( 2 pi ). Evaluating with a calculator we get x = { -5.62, -1.13 }. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have the same answer but I found this one difficult to do ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: * 1.2.106 \ 98 (was 1.3.90). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) box vol 4 ft^3 by cutting 1 ft sq from corners of rectangle the L/W = 2/1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I didnt even understand how to start this one. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I dont understand how you got the equation 1(x-2)(2x 2) = 4, I understand the 4 but not the left side of the equation. Boom it just hit me while I was typing the last sentence I wasnt thinking x is the shorter side 2x is the longer side and the minus 2 is the taking away of each corner. I understand what I should have done to successfully complete this problem. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 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: Solve for 15 m -4.9t^2 + 20t 15 = 0 plug into quadratic fomula x = [-b +- sqrt (b^2 4ac)] / 2a [-20 +- sqrt(20^2 4*-4.9*-15)] / (2 * -4.9) = there will be two values t = .99 and t = 3.09 on the way up and on the way down Solve for 0 meters -4.9t^2 + 20t = 0 plug into quadratic fomula x = [-b +- sqrt (b^2 4ac)] / 2a [-20 +- sqrt(20^2 4*-4.9*0)] / (2 * -4.9) = [-20 +- sqrt(20^2) / -9.8 = two values t= 4.08 and t= 0 from the beginning to the end Solve for 100m -4.9t^2 + 20t 100 = 0 plug into quadratic fomula x = [-b +- sqrt (b^2 4ac)] / 2a [-20 +- sqrt(20^2 4*-4.9*-100)] / (2 * -4.9) = We come up with a negative number and find that the object will not reach 100 m confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. The only one that I had a problem with was the volume of the box but I see what I missed the first time I looked at it. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. The only one that I had a problem with was the volume of the box but I see what I missed the first time I looked at it. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!