course Mth 158 Sorry, about the last few problems but I think I understand the quadatic equations as long as they are not in problem solving form. But I have been having problems with medications for/from dental surgery, allergic reactions and have slept for the past 3 days. Sorry some of my work is late. But I promise I will catch up. Thanks {í£ÈâzE€Ize½Ôš×xå·æ±ŒS¥™Íá¹ìassignment #012
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11:10:00 1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.
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RESPONSE --> z^ - z - 6 =0 Solve by factoring (z - 3)(z + 2)=0 z - 3 =0 or z + 2 =0 z=3 or z= -2 {-2,3} solution set check by using solution set {-2,3} in place of z (-2)^2-(-2)-6=0 (3)^2-3-6=0 4+2-6=0 9-3-6=0 6-6=0 0=0 0=0 check by solving the factor (z - 3)(z + 2)= z^2-3z+2z-3= z^2-1z-6= z^2-z-6= confidence assessment: 3
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11:10:39 ** 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, but note that -2 also checks: (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0. **
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RESPONSE --> whoooah! I got it right....lol self critique assessment: 2
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11:33:59 1.1.72. Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.
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RESPONSE --> Solve by factoring, but first I must get the equation into standard quadratic form. ax^2+bx+c=0 x^3+6x^2-7x=0 factor out the x x(x^2+6x-7)=0 then divide to get rid of the x x^2+6x-7=0 I think? (x+7)(x-1)=0 x+7=0 or x-1=0 x= -7 or x=1 solution set {-7,1) check (-7)^3+6(-7)^2-7(-7)=0 (1)^3+6(1)^2-7(1)=0 -343+294-(-49)=0 1+6-7=0 0=0 0=0 confidence assessment: 2
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11:34:33 ** 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. **
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RESPONSE --> YES! I am getting hang of this. self critique assessment: 2
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11:38:26 1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.
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RESPONSE --> Ok? (there is no equation so I will explain how I solved the one before this question) First I needed to get rid of an x to get the equation into standard quadratic form: ax^2+bx+c=0. Then I solved by factor the equation as (x+7)(x-1)=0. Thus giving me a solution set of { -7, 1}. confidence assessment: 3
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11:39:45 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}
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RESPONSE --> ok I thought something was up here I didn't see an equation in the question before, and there wasn't a scroll down bar....soooo....sorry! self critique assessment: 2
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11:49:51 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.
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RESPONSE --> x(x+4)=12 Ok so I factor in the x and put into x^2+4x=12 standard quadrtic form x^2+4x-12=0 by subtracting 12 from both sides (x+6)(x - 2)=0 Now I factor out the equation x+6=0 or x - 2=0 x= -6 or x=2 the solution set { -6,2} check (-6)^2+4(-6)-12=0 36-24-12=0 -2=0 doesn't check unless -6(-6+4)=12 12=12 confidence assessment: 2
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11:50:34 ** 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 --> ok I got this one and checking the solution set was a little confusing??? self critique assessment: 2
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12:08:08 1.2.38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.
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RESPONSE --> Solve by factoring I need to get this equation into standard quadratic form, to do so: x+12/x=7 to get rid of the fraction by multiply by x(x+12/x)=7(x) x to both sides x^2+12=7x needs to be in standard quadratic x^2-7x+12=0 form by subtracting both sides by 7x (x - 3)(x - 4)=0 x - 3=0 or x - 4=0 x=3 or x=4 solution set {3,4} confidence assessment: 2
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12:12:16 ** 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 --> YEAH! self critique assessment: 2
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12:15:21 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 --> Solve by the square method. (x+2)^2=1 x+2=+ or - sqrt1 x+2=1 or x+2= -1 x= -1 or x= -3 solution set {-3,-1} confidence assessment: 2
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12:15:59 ** (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 --> yaaay! lol I like knowong that I understand this. self critique assessment: 2
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12:54:39 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 --> Solve by completing the square x^2+mx+(m/2)^2=(x+m/2)^2 x^2+2/3x-1/3=0 get rid of fractions by multiply both 3x^2+2x-1=0 sides by 3 then add 1 to both sides 3x^2+2x+(2/2)^2=1 3x^2+2x+2=1 now subtract1 from both sides 3x^2+2x+1=0 factor out the equation (3x-1)(x+1)=0 zero property factor 3x-1=0 or x+1=0 x=1/3 or x= -1 solution set {-1,1/3} confidence assessment: 3
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12:54:57 ** 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 self critique assessment: 2
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13:04:21 1.2.52 (was 1.3.42). Explain how you solved the equation x^2 + 6x + 1 = 0 using the quadratic formula.
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RESPONSE --> Solve by Quadratic formula ax^2+bx+c=0 x=(-b+or-sqrtb^2-4ac)/2a x^2+6x+1=0 a=1,b=6,c=1 b^2-4ac=(6)^2-4(1)(1)=36-4=32 x= [-(6)|+or-|sqrt32]/2(1) x=(-6|+or-|sqrt32)/2 I'm guessing the solution set is {(-6-sqrt32)/2,(-6+sqrt32)/2} confidence assessment: 1
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13:06:09 ** 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 --> ok I got the first solution set without simplifying them. Its good that I did understand that was about as far as I could go. self critique assessment: 2
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13:12:41 1.2.72 (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 --> ok this one is a little more confusing, bu however I get this.... {-(15sqrt(2))|+or-|sqrt198.8}/6.24 confidence assessment: 0
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13:21:21 ** 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)) ± 198.68] / ( 2 pi ). Evaluating with a calculator we get x = { 5.62, 1.13 }. DER**
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RESPONSE --> ok so the sqrt cancelled out the (-15sqrt2)^2 so you leave pi as is and not change to 3.14 thats where I had problems. And where is pi on my calculator I am going to look into that. Sorry...lol I am silly sometimes. self critique assessment: 2
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13:28:48 1.2.98 (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 --> Let x represent the length of the side (x-2)(x-2)(1)=4 x^2-4x+4+4 x^2-4x=0 x(x-4)=0 x=0 or x=4 x can not =0 confidence assessment: 1
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13:33:47 ** 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 --> ok so did that completely wrong. I will work on more problems like this one. sorry self critique assessment: 2
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13:43:51 1.2.100 (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 --> Ok I am very sorry but I am lost when it comes to problem solving. confidence assessment: 0
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13:50:23 ** 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 --> as simple as that, ok so I feel a little stupid for not seeing it to be that easy. self critique assessment: 2
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13:51:14 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The last couple of problems got me flustered but other than that I felt I did ok. confidence assessment: 2
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