course MTH 158
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20:59:58 query 1.2.5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.
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RESPONSE --> since the equation is in standard form you factor the left side to get (z-3)(z+2)=0 now set each factor equal to 0 now you have z-3=0 to give you 3 and z+2=0 to get -2 so your solution set is {3, -2}
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21:00:16 ** 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 --> Ok
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21:03:26 **** query 1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.
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RESPONSE --> v^2+7v+6=0 first i factored out the left side to get (v+6)(v+1)=0 now set each factor to 0 to get (v+6=0 and you get -6 which checks out and then you get v+1=0 to give you -1 which also checks out.
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21:03:35 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
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21:08:53 **** 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 --> first i multiplied what was in () by x to give me x^2+4x=12, now subtract the 12 over to get x^2+4x-12=0 now factor to get (x+6)(x-2)=0 set each factor to 0. x+6=0 x=-6 which checks out and then x-2=0 x=2 which also checks out so {-6,2}
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21:08:59 ** 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
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21:13:21 **** query 1.2.38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.
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RESPONSE --> multiply by x to get x^2+12x=7x subtract the 7x now you have x^2+12x-7x=0 add the like terms x^2+5x=0
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21:13:52 ** 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 --> I multiplied the 12 by x too
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21:17:04 **** 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 --> you use the square rooth mehto to get x+2=plus or minus sqrt1 x+2=+-1 x+2=1 and x+2= -1 x=-1 and x=-3 {-1, -3}
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21:17:08 ** (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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22:53:21 **** 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 --> multiply by 3 to get 3x^2+2x-1x=0 then you rewrite the equation so its now 3x^2+2x=1 now divide my 3 so now its x^2+2/3x=1/3 now add 2/3 to each side x^2+2/3x+2/3=1/3+2/3 and i'm lost after this part...
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22:53:53 ** 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 --> i should have just factored it after i multiplyed by 3
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22:58:31 **** query 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 --> use the formula x=-b+-sqrtb^2-4ac/2a and fill in the numbers a=1 b=6 c=1 x=-6+-sqrt6^2-4*1*1/2*1 x=-6+-sqrt(36-4)/2 x=-6+sqrt(32)/2 or -6-sqrt(32)/2
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22:59:11 ** 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
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23:00:19 **** query 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 --> i still don't have my Ti83 but you do the same, fill in the numbers so that a=pi b=15sqrt(2) and c=20 with a Ti83 you can fill in the exact equation because it has those symbols.
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23:00:22 ** 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 --> ok
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23:01:16 **** query 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 --> ?
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23:01:20 ** 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
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23:01:47 **** query 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 --> ?
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23:01:52 ** 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 --> ok
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23:02:06 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> jsut had a few problems with the word problem set
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