Assign 11

course Mth 158

10-5-09 5:35

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

011. `* 11

* 1.2.13 \ 5. Explain, step by step, how you solved the equation z^2 - z - 6 = 0 using factoring.

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Your solution:

z^2 - z - 6 = 0 factor the left side of the equation

(z + 2) ( z - 3) = 0 Which gives us the set

{-2, 3}

confidence rating: 3

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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. **

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Self-critique (if necessary):

There are no discrepancies.

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Self-critique Rating: 3

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Question: * 1.2.14 (was 1.3.6). Explain how you solved the equation v^2+7v+6=0 by factoring.

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Your solution:

v^2+7v+6=0 factor the left side of the equation

(v + 1) (v + 6) = 0 Leaving us with the set

v = {-1, -6}

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}

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Self-critique (if necessary):

There were no discrepancies.

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Self-critique Rating: 3

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Question: * 1.2.20 (was 1.3.12). Explain how you solved the equation x(x+4)=12 by factoring.

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Your solution:

x(x+4) = 12 factor the left side of the equation and subtract 12 from both sides

x^2 + 4x – 12 = 0 to get the standard form, then factor the left

(x-2)(x+6) = 0 to give us the set

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} **

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Self-critique (if necessary):

There are no discrepancies.

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Self-critique Rating: 3

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Question: * 1.2.26 \ 38 (was 1.3.18). Explain how you solved the equation x + 12/x = 7 by factoring.

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Your solution:

x + 12/x = 7 by dividing both sides by the denominator x

x^2 + 12 = 7x then subtract 7x from both sides to get the standard form

x^2 -7x + 12 = 0

(x – 3)(x – 4) = 0 to get the set

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} **

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Self-critique (if necessary):

No discrepancies.

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Self-critique Rating: 3

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Question: * 1.2.32 (was 1.3.24). Explain how you solved the equation x + 12/x = 7 by the square root method.

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Your solution:

x + 12/x = 7

x + 12 = ± sqrt7

I’m not sure about this one. The 12/x threw me off. I couldn’t follow it in the book to solve.

confidence rating: 0

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Given Solution:

The problem as I believe it appears in the query is (x + 2)^2 = 1.

I don't think the file has changed recently (the date showing on the Web is 3/1/09).

You apparently got an incorrect document, but I haven't yet figured out how this happened. I haven't seen this problem before, and hopefully it won't happen again.

The solution below is correct for the equation (x + 2)^2 = 1.

(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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Self-critique (if necessary):

I have no idea what the solution is saying about the problem. Where is the 12/x?

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Self-critique Rating:

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Question: * 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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Your solution:

x^2 + 2/3 x - 1/3 = 0 First you multiply both sides by 3 to get

3x^2 + 2x – 1 = 0 then factor the left

(3x – 1)(x + 1) = 0 to get the set

x = {1/3, -1} (I really don’t see how it comes up with 1/3, but that’s how the book explains it.)

confidence rating: 1

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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.

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Self-critique (if necessary):

The only thing that confuses me is the 1/3. Is that because of the 3x?

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.

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Self-critique Rating: 1

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Question: * 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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Your solution:

x^2 + 6x + 1 = 0

Using the quadratic formula ax^2 + bx + c = 0 for this equation you get

a = 1, b = 6, and c = 1

Then evaluate the discriminant b^2 – 4ac

6^2 – 4(1)(1) =

36 – 4 = 32 so there is 2 real solutions

x = 6 ± sqrt32 / 2

That’s about as far as I can follow along in the book. I get confused how to go further with the solution.

That should be written

x = (-6 ± sqrt32) / 2.

The only thing missing once you have this is the simplification, which results from the fact that

sqrt(32) = sqrt(16 * 2) = sqrt(16) * sqrt(2) = 4 * sqrt(2).

confidence rating: 0

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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) }. **

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Self-critique (if necessary):

I’m still confused.

You have most of this. You would therefore be able to tell me pretty clearly what you do and do not understand.

See my notes, inserted into your solution.

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Self-critique Rating: 0

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Question: * 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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Your solution:

pi x^2 + 15 sqrt(2) x + 20 = 0 for this equation,

a = pi, b = 15 sqrt(2), c = 20

Then evaluate the discriminant b^2 – 4ac

(15 sqrt(2))^2 – 4(pi)(20) using my calculator, I came up with

450 – 80pi = 370pi

450 pi - 80 pi = 370 pi, but

450 - 80 pi = 450 - 250 = 200, approx., as you have below

450 – 251.2 = 198.8 (which is a repeated solution)

x = (-1.5 sqrt(2) ± sqrt198.8 / 2

This would be

x = (-15 sqrt(2) ± sqrt198.8 ) / 2.

Evaluate

(-15 sqrt(2) + sqrt198.8 ) / 2 and

(-15 sqrt(2) - sqrt198.8 ) / 2

and then I get lost here.

confidence rating: 1

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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**

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Self-critique (if necessary):

I just can’t seem to follow along after the place I stopped.

see my note

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Self-critique Rating: 0

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Question: * 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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Your solution:

Cut out 1 sq ft from corners

Length = 2x – 2ft

Width = x – 2ft

This gives the ratio of 2/1

Then the formula is h*w*l, we know the height is 1ft from the sq ft cut out on the corners and volume = 4ft^3

1 (x-2)(2x-2) = 4ft^3 factoring the left

2x^2 -6x +4 = 4

x^2 -3x +2 = 2

(x-2)(x-1) = 2 subtract 2 from both sides to get 0 property

(x-2)(x-1) -2 = 0

x^2 – 3x = 0 factor the left

x(x-3) = 0 leaving us with the set

x = {0, 3}

confidence rating: 1

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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.

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Self-critique (if necessary):

I really don’t know if I did this right, but the solution looks the same. I did insert a drawing so I could see what I was doing. I don’t know if it will translate to the submission form the correct way or not, so if it looks funny that’s why.

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Self-critique Rating:

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Question: * 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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Your solution:

This one is really confusing. I tried to set it up using the formula like this:

x = -4.9t^2 + 20t where s = 15m high

-4.9t^2 + 20t = 15 subtracting 15 from both sides to give us a zero product

-4.9t^2 + 20t – 15 = 0

Here is where I just couldn’t follow along.

t = {-20 ± sqrt [20^2 – 4(-4.9)(-15)} / 2

evaluate

20^2 – 4(-4.9)(-15), and use this to evaluate

sqrt(20^2 – 4(-4.9)(-15)). Then use this result to find

-20 + sqrt(20^2 – 4(-4.9)(-15)) and

-20 - sqrt(20^2 – 4(-4.9)(-15))

Divide both of these by 2 and you'll have it.

confidence rating: 0

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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. **

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Self-critique (if necessary):

I still don’t understand it.

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Self-critique Rating:

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Question: * Add comments on any surprises or insights you experienced as a result of this assignment.

There was one thing that I realized, and it’s funny that I can be 46 years old and never knew this or understood why they call it a quadratic equation. When I was looking at the picture in the text book and there were 4 areas to be solved, it hit me like a ton of bricks!!

I am feeling like I have taken on more than I can do with this type of math. I struggle so hard with understanding it. It has taken me the good part of the weekend to read and read to try and understand just bits and pieces of it. I really don’t know what to do, but I know that I am giving it 100%.

I've inserted a number of notes. You might well have additional questions. If you understand rything fine, but if you have questions do as in the following instruction.

&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#