#$&* course Mth 158 9/22 4 011. `* 11
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, because sometimes we get extraneous roots. But note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v^2+7v+6=0 factor (v+6)(v+1)= 0 v+6=0 and v+1=0 v=-6 and v=-1 {-6,-1} both work in the equation confidence rating #$&*: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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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 distribute x^2+4x=12 subtract 12 from both sides x^2+4x-12=0 factor (x+6)(x-2)=0 x+6=0 and x-2=0 x=-6 and x=2 {-6,2} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x + 12/x = 7 multiply both sides by x x^2+12=7x subtract 7x on both sides x^2-7x+12=0 factor (x-3)(x-4)=0 x-3=0 and x-4=0 x=3 and x=4 {3,4} confidence rating #$&*: 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: ok ********************************************* 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 distribute (x+2)(x+2)=1 distribute some more x^2+2x+2x+4=1 x^2+4x+4=1 subtract 1 from both sides x^2+4x-3=0 factor (x+3)(x+1)=0 x+3=0 and x+1=0 x=-3 and x=-1 {-3,-1} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * 1.2.44 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ok this one has been messing with me, but I know you start with x^2 + 2/3x - 1/3 = 0 and that you multiply both sides of the equation by 3 so you should get 3x^2 + (6/3)x - 3/3= 0 which simplifies to 3x^2+2x-1=0 factor (3x-1)(x+1)=0 3x-1=0 or x+1=0 3x=1= 1/3 or x=-1 {1/3, -1} confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: quadratic formula is x= [-b(+-) sqrt(b^2-4ac)]/2a and since the equation is ax^2+bx+c=0 so with x^2 + 6x + 1 = 0 a=1, b=6, c=1 so the quadratic formula reads x= [-6(+-) sqrt(6^2-4*1*1)]/2*1 x= [-6(+-) sqrt(36-4)]/2 x= [-6(+-) sqrt(32)]/2 x= [-6(+-) sqrt(16*2)]/2 x= [-6(+-) 4sqrt(2)]/2 x= [-3(+-) 2sqrt(2)] so x= [-3+ 2sqrt(2)] or x= [-3- 2sqrt(2)] confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^2 + 6x + 1 = 0 we identify our equation as a quadratic equation, having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ± sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ± sqrt(36 - 4) / 2 x = { -6 ± sqrt (32) ] / 2 36 - 4 = 32, so x has 2 real solutions, x = [-6 + sqrt(32) ] / 2 and x = [-6 - sqrt(32) ] / 2 Our solution set is therefore { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Now sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * 1.2.78 \ 72 (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a=pi, b= 15sqrt(2,) c=20 x= [-15sqrt(2) (+-) sqrt(15sqrt(2) - 4pi20)]/2pi which simplifies to x= [-15sqrt(2) (+-) sqrt((225*2) -( 80*3.14))]/2pi x= [-15sqrt(2) (+-) sqrt(450 - 251.2]/2pi x= [-15sqrt(2) (+-) sqrt(198.8)]/2pi calculator says x={-1.133, -5.622} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: For visual purposes I drew a rectangle and a little square out of each corner. If l/w=2/1 then the length is twice as long as the width. So if the width of the rectangle = x then the length= 2x . Now note the 1ft squares being cut out of each corner, and you see that you'll have to subtract 2ft from each side. So now width would = x-2 and length=2x-2 and when we fold up the remaining sides of the rectangle we see that the height is still only 1ft. The volume is already given as 4ft^3 so we know length=2x-2 width= x-2 height=1 volume=4 and we can put the equation together as (2x-2)(x-2)1=4 distribute to get 2x^2-6x+4=4 divide both sides by 2 x^2-3x+2=2 subtract 2 from both sides and get x^2-3x=0 you can simplify that to x(x-3)=0 and get x=0 or x-3=0 so x=0 or x=3 , but for this situation x=3 is the only one that makes sense. So going back to what we originally said about the rectangle where length=2x-2 and width= x-2 we can now plug in 3 and get length=2*3-2=4 and width=3-2=1 therefore we know the dimensions of the box are 4ft x 1ft x 1ft= 4ft^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: (a) 15=-4.9t^2 +20t 0 = -4.9t^2+20t-15 so a= -4.9, b=20, and c= -15 t=(-20(+-) sqrt(20^2- 4*-4.9*-15))/(2*-4.9) t= (-20(+-) sqrt(400- 294))/(-9.8) t=(-20(+-) sqrt(106))/(-9.8) use calculator and find that t approximately = {.99, 3.09} so the object will be at 15m height at about .99 seconds and 3.09 seconds (b) 0=-4.9t^2 +20t 0= t(4.9t +20) so t= 0 or... -4.9t+20=0 -4.9t= - 20 t= 4.081 so t={4.081, 0} (c) 100=-4.9t^2 +20t 0= -4.9t^2 +20t - 100 so a=-4.9, b=20, and c=-100 t=(-20(+-) sqrt(20^2- 4*-4.9*-100))/(2*-4.9) t= (-20(+-) sqrt(400-1960))/(-9.8) t= (-20(+-) sqrt(-1560))/(-9.8) no solution, so I guess the object can never make it to 100ft. Too bad. 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: ok ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!