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20:07:17 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 --> factor (z^2 - z -6) and set each factor to 0
(z-3)(z+2)=0 z-3 =0 or z + 2 =0 z=3 z = 2 {2,3}.................................................
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20:07:46 ** 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. **......!!!!!!!!...................................
RESPONSE --> right
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20:09:23 **** query 1.2.14 (was 1.3.6). Explain how you solved the equation by factoring.
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RESPONSE --> ok there isn't a problem typed in here so I will do #14 from the book:
v^2 + 7v + 6 = 0 (v+6)(v+1) = 0 v=-6 v = -1 {-6,-1}.................................................
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20:09:36 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}......!!!!!!!!...................................
RESPONSE --> right
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20:11:15 **** 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 subtract 12 from both sides
x (x+4) - 12 = 0 multiply out: x^2 + 4x - 12 = 0 (x+6)(x-2) = 0 {-6,2}.................................................
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20:11:33 ** 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} **......!!!!!!!!...................................
RESPONSE --> right
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20:15:30 **** 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 --> first mulitply each side by x to get rid of fraction =x^2 + 12 = 7x subtract 7x from both sides
= x^2 - 7x + 12 = 0 (x-4)(x-3)=0 {3,4}.................................................
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20:15:43 ** 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} **......!!!!!!!!...................................
RESPONSE --> right
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20:18:26 **** 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 = + - sqrt1 x + 2 = +- 1 x ={ -1,-3}.................................................
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20:18:56 ** (x + 2)^2 = 1 so that
x + 2 = ?sqrt(1) giving us x + 2 = 1 or x + 2 = -1 so that x = {-1, -3} **......!!!!!!!!...................................
RESPONSE --> right
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20:33:31 **** 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 --> half of (b) squared will be added to each side then solve with sqrt root method
x^2 + 2/3x - 1/3 = 0 x^2 + 2/3x = 1/3 x^2 + 2/3x + (4/3)^2 = 1/3 + (4/3)^2 (x + 4/3)^2 = 19/9 x + 4/3 = + - sqrt (19/9) x + 4/3 = +- 19/9 x = -4/3 + 19/9 x = -4/3 - 19/9 x=7/9 x = -31/9.................................................
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20:35:08 ** 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**......!!!!!!!!...................................
RESPONSE --> I thought this one had to be done by completing the square. I see what you have done by multiplying by 3 and factoring. This is much easier.
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20:42:09 **** 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 --> enter the corresponding values in for a, b and c into the formula
x= [ -(6) + - sqrt [(6)^2 -4 *1 *1] ] /2*1 x = [-6 +- sqrt (32) ] / 2 x =( -6 + - 8 )/2 {-7, 1 }.................................................
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20:44:32 ** 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) }. **......!!!!!!!!...................................
RESPONSE --> When working the problem I solved the sqrt of 32 as sqrt( 8*4) =2^3 = 8. Is my answer incorrect?
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20:54:57 **** 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 --> a = pi, b = 15 sqrt(2) c = 20
x= [-15sqrt(2) + - sqrt ( (15sqrt2)^2 - 4 *pi * 20]/2*pi [-21.2 + - sqrt(450 - 251.3}]/6.28 ]-21.2 +- sqrt (198.7)]/6.28 [-21.2 +-14.1]/6.28 x = {-1.13, 1.13}.................................................
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21:00:23 ** 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**......!!!!!!!!...................................
RESPONSE --> So, we don't muliply out -15sqrt(2) until the last step. I don't see how you got one answer the same as mine and another different when our steps were almost the same.
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21:08:15 **** 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 --> After working number 97 and trying to go by the same reasoning I found there, i came up with this solution:
since l = 2w let w=x and l=2x 1ft is the size being cut so that is the height and equation must equal 4 9(f^3) (x - 2)(2x-2)1 = 4 2x^2 - 6x + 4 = 4 2x^2 - 6x = 0 2x(x -3) = 0 2x=0 x -3 = 0 since x cannot equal 0 in this problem the answer is 3. 3ft X 3ft.................................................
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21:11:36 ** 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. **......!!!!!!!!...................................
RESPONSE --> I see I have done my differently than yours again but gain the same answer. Is either way correct?
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21:33:05 **** 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 --> a) let equation equal 15 m since that is what we are measuring. -4.9t^2 + 20t = 15 -4.9t^2 + 20t - 15 = 0 solve with quadratic form. [-20 +- sqrt( (20)^2 + 4 * -4.9 * -15]/2*-4.9
= [-20 +- sqrt (694)]/-9.8 [-20 +- 26.3]/-9.8 x = -0.64 x = 4.7 since a neg. number doesn't make sense, 4.7 seconds is the answer. b)ground is zero so set equation to zero and solve -4.9t^2 + 20t = 0 [-20 +- sqrt( (20)^2 - 4 * -4.9 * 0] / 2*-4.9 [20 +- sqrt (400)]/-9.8 (20 +- 20) / -9.8 -40/-9.8 = 4.08 the other answer is neg. so the object will reach ground in 4.08 sec. c)for this question I found a formula in another math textbook I have that I think applies to this problem. height=-b/2a substituting the number from the prob. gives: -20/-19.8 = -1.01 then plugging in that number to our equation for s: s = -4.9 * -1.01 + 20 * -1.01 = -15.25 No, it will not reach 100m.................................................
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21:39:17 ** 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. **......!!!!!!!!...................................
RESPONSE --> I double checked my math on part a and still came up with the same answers so I am unsure how I miscalculated.
Your answer for part c was much less complicated than what I attempted to do. I will use that next time..................................................
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21:40:05 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found the word problems in this section to be pretty hard. They were quite time consuming for me.
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