#$&* course Mth 158 2-28 12:30am f011. `* 11
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I factored this and came up with (z + 2)(z - 3) = 0 Which broke down to z + 2 = 0 and z - 3 = 0 This gave me the set {-2, 3} -2 however, doesn't check out, but only 3 does, so the solution is: z = 3 INSTRUCTOR COMMENT: It's good that you're checking out the solutions, because sometimes we get extraneous roots. But note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Factor V^2 + v + 6v + 6 = 0 V(v + 1) + 6(v + 1) = 0 (v + 6)(v + 1) = 0 V= -6, -1 confidence rating #$&*: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 and 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: First subtract 12 from both sides to make the right equal to zero X^2 + 4x - 12 = 0 Factor X^2 + 6x -2x -12 = 0 X(x + 6) - 2(x + 6) = 0 (x - 2)(x + 6) = 0 X = 2, -6 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x(x+4)=12 apply the Distributive Law to the left-hand side: x^2 + 4x = 12 add -12 to both sides: x^2 + 4x -12 = 0 factor: (x - 2)(x + 6) = 0 apply the zero property: (x - 2) = 0 or (x + 6) = 0 so that x = {2 , -6} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First multiply both sides by x to get rid of the fraction X^2 - 7x + 12 = 0 Factor X^2 - 3x - 4x + 12 = 0 X(x - 3) -4 (x - 3) = 0 (x - 3)(x - 4) = 0 X= 3, 4 confidence rating #$&*: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: The square root of (x + 2)^2 is x + 2, and the square root of 1 is 1 So we can reduce the equation down to X + 2 = 1 or x + 2 = -1 Then subtract 2 from both sides X = -1 or X = -3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.38 (was 1.3.36). Explain how you solved the equation x^2 + 2/3 x - 1/3 = 0 by completing the square. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First multiply both sides by 3 to get rid of the fractions 3x^2 + 2x - 1 = 0 Factor 3x^2 + 3x -x - 1 = 0 3x(x + 1) - 1(x + 1) = 0 (3x - 1)(x + 1) = 0 X = {1/3, -1) confidence rating #$&*: 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:3 ********************************************* 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: Plug the a, b, and c values into the equation [-6 ± sqrt(6^2 - 4)]/2 Simplify [-6 ± sqrt(32)]/2 [-6 ± 4*sqrt(2)]/2 divide the components of expression by 2 to further simplify the solution set is as follows { [-3 + 2*sqrt(2)], [-3 - 2*sqrt(2)] } confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^2 + 6x + 1 = 0 we identify our equation as a quadratic equation, having the form a x^2 + b x + c = 0 with a = 1, b = 6 and c = 1. We plug values into quadratic formula to get x = [-6 ± sqrt(6^2 - 4 * 1 * 1) ] / 2 *1 x = [ -6 ± sqrt(36 - 4) / 2 x = { -6 ± sqrt (32) ] / 2 36 - 4 = 32, so x has 2 real solutions, x = [-6 + sqrt(32) ] / 2 and x = [-6 - sqrt(32) ] / 2 Our solution set is therefore { [-6 + sqrt(32) ] / 2 , [-6 - sqrt(32) ] / 2 } Now sqrt(32) simplifies to sqrt(16) * sqrt(2) = 4 sqrt(2) so this solution set can be written { [-6 + 4 sqrt(2) ] / 2 , [-6 - 4 sqrt(2) ] / 2 }, and this can be simplified by dividing numerators by 2: { -3 + 2 sqrt(2), -3 - 2 sqrt(2) }. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 1.2.69 \ 1.2.78 \ 72 (was 1.3.66). Explain how you solved the equation pi x^2 + 15 sqrt(2) x + 20 = 0 using the quadratic formula and your calculator. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Applying the quadratic formula, we get [(-15*sqrt(2))± sqrt(15*sqrt(2)^2 - 4(pi)(20))]/2pi [(-15*sqrt(2))± sqrt(15*sqrt(2)^2 - 80pi)]/2pi Using a calculator, we get (-21.21 ± 5.36)/6.28 X = -2.52, or -4.23 confidence rating #$&*: 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’m not sure at all about this problem…I don’t have a scientific calculator so I just use my regular one. I followed the order of operations for this problem but could not come up with the same answers that were provided. Not sure what I was doing wrong. ------------------------------------------------ Self-critique Rating:2
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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:3 ********************************************* 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: To find the time at which an object is 15m off the ground, we set the height at 15 for s So the equation would read 15 = -4.9t^2 + 20t Subtract 15 from both sides 0 = -4.9t^2 + 20t - 15 Using the quadratic equation, we get [-20 ± sqrt(20^2 - 4(-4.9)(-15)]/2(-4.9) This simplifies to X = 4.02 and x = .06 Which means that the object would pass through the 15m mark going up at .06 and on the way down at 4.02 To find when the object hits the ground, set s to 0 0 = -4.9t^2 + 20t Use the quadratic equation to solve [-20 ± sqrt(20^2 - 4(-4.9)]/2(-4.9) So t= 0 and 4.1 So the object leaves the ground at 0m and then hits the ground again at 4.1m For 100m, plug 100 in for s 100 = -4.9t^2 + 20t Subtract 100 from both sides 0 = -4.9t^2 + 20t - 100 Use the quadratic equation to solve for t [-20 ± sqrt(20^2 - 4(-4.9)(-100)]/2(-4.9) The discriminant is negative, so the object will not rise to 100m confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION I was lost on this question and even reading the solution, Im still confused about it. Do you have any suggestions on how to look at it in a different way??? INSTRUCTOR RESPONSE s = -4.9 t^2 + 20 t means that if you plug in a value of t, you get the height, which is represented by the variable s. Also if you want to find the value of t that gives you a certain height, you plug in that height for s and solve the equation for t. The first question asks you to find when the height is 15 meters. So you plug in 15 for s. What equation do you get? You get the equation 15 = -4.9 t^2 + 20 t. Now you solve the equation for t. How do you do that? The equation is quadratic, since it contains both t^2 and t. The standard form for a quadratic equation is a t^2 + b t + c = 0 In this form you can try to factor the left-hand side. If this is possible you can then apply the zero property, as you've done in some of the preceding problems. If you can't figure out how to factor the equation (and in real-world problems you usually can't), you can use the quadratic formula. In this case you rearrange the equation 15 = -4.9 t^2 + 20 t to the form -4.9 t^2 + 20 t - 15 = 0 and pretty quickly realize that you won't be able to factor it. So you use the quadratic formula, as shown in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have always been horrible at word problems. I understand the concept of the heights and what needs to be plugged in in this particular one, but when it comes to getting the exact decimal values, I always get it a few points off and I’m not sure how. I do understand the concept and I have the quadratic equation memorized. ------------------------------------------------ Self-critique Rating:2