course Mth 158 I titled the last assignment assignment 11 011. `query 11 College Algebra 06-21-2008
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11:19:13 **** query 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.
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RESPONSE --> I mistook the + for a -.Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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11:21:43 ** 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} **
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RESPONSE --> x(x+4) = 12 x^2 + 4x = 12 x^2 + 4x -12= 12-12 (x-2)(x+6) = 0 x-2= 0 or x+6 =0 x= 2 or x=-6 (2, -6
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11:24:22 **** query 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.
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RESPONSE --> x + 12/x =7 Multiply each side by x leaving (^2 =12 = 7x Factor (x-4) (x-3) x-4=0 or x-3=0 x= 4 or x=3 Solution set is ( 3,4)
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11:24:34 ** 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} **
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RESPONSE --> ok
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11:28:37 **** query 1.2.32 (was 1.3.24). Explain how you solved the equation (x+2)^2 = 1 by the square root method.
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RESPONSE --> (x+2)^2=1 x+2 = sq rt 1 x+2= +- 1 x+2 = 1 or x+2 = -1x= -1 or x=-3 Solution set is (-1, -3 )
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11:28:43 ** (x + 2)^2 = 1 so that x + 2 = ± sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} **
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RESPONSE --> ok
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11:40:32 **** query 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.
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RESPONSE --> x^2 + 2/3 x -1/3= 0 Add 1/3 to each side x^2 =2/3x = 0 +1/3 Complete the square by adding 1/9 x^2 + 2/3x +1/9 = 4/9 (x - 1/3 )^2 = 4/9 x-1/3 = +- sqrt 1/9 x= 1/3 or x= -1/9
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11:42:10 ** 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. DER**
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RESPONSE --> Ok
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12:06:41 **** query 1.2.50 \ 52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula.
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RESPONSE --> x^2 + 6x +1 = 0 a= 1 b=6 c=1 x= -6 +- sqrt of- 6^2 - 4 (-1)(1)/ 2 (1) x= -6 +- sqrt 36 - 4/2 = -6 +- 6-4 /2 x=-6 + 2/2 or x=-6 -2/2 x= -3/2 or x=-8/2 Solution set is (-3/2 or -8/2)
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12:12:32 ** Starting with x^2 + 6x + 1 = 0 we identify this as 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 , [-6 - sqrt(32) ] / 2 } Note that 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) }. **
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RESPONSE --> It was hard to follow this but I can see that the sq rt of 32 can be simplified.
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12:27:31 **** query 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.
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RESPONSE --> Pi x^2 - 15 sqrt 2x +20 = 0 a= pi b= -15 c= 20 x= -(-15) + or - sqrt (-15)^2 - 4 pi (20)/2 pi =-(-15) + or - 225- 4 (3.14) (20)/ 2 pi x= 15 + or - 80 pi/2 pi x= -65 pi/2pi or x= 95 pi/ 2 pi
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12:28:52 ** 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 ). (-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**
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RESPONSE --> Using the calculator you mutiply out you answer which I didn't do
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12:49:22 **** query 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.
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RESPONSE --> x= lenght of side length is 2(x+2) Width is x-2 height is 1 ft = l *w*h 64 = (2x+2) (x-2) (1) 64= 2x^2 -2x - 4
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12:54:12 ** 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. (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 checks out fine. **
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RESPONSE --> For the rectangle you have 2x-2 Distribute and then solve.
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13:25:23 **** query 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.
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RESPONSE --> -4.9t^2 + 20 t + 15 =0 t^2 - 4t - 3 =0 (t-1) (t-3) = 0 t= 3 seconds
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13:32:14 ** 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. 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. 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. We conclude that this object will not rise 100 ft. **
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RESPONSE --> I see you use the quadratic formula to solve the problem after you determine your equation
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