Help with college algebra final review...

[MegaVovaN]

Hi all!
First of all, this is not homework, it is a review for Final exam, I do not get extra credits for completing it (the review), or anything like that.

My coursecompass.com account expired today, and I can't use it for studying

I have 60 problems on review, and I just did 30 (used 12 sheets of printer paper for scratching). On couple of them, my answer != answer in the answer key. Some explanations and help will be appreciated :thumbsup:

#7)

3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:







#9)

(x+28)/(x+1) < 4
I can understand x != -1, but how did they get x !=8? What were the steps?





#19)
Find slope and y-intercept.
x-5y - 5 = 0
This is probably so simple it's humiliating. What are the STEPS to finding slope and y-int? I do not need the answer, it's already given in answer key (slope = 1/5 and y-int (0,-1)).




#20)
Write the equation of the line in standard form.
x-int is -5
y-int is 3
I know std form is Ax+By = C, but I do not understand how to go about the steps? Answer is (3x-5y = -15).



#29)
The base and height of a triangle must have a sum of 40 inches. Find the base and height of the triangle whose area is as large as possible.

This is probably also dead simple, but I am stuck. I know that
b+h = 40
Area = 0.5 * bh

Answer is base 20 in, height 20 in.



Here's the PDF with review:
Text

Thanks for your help, o allmighty math gurus! And may your respect++

 

[hypn0tik]

Originally posted by: MegaVovaN
Hi all!
First of all, this is not homework, it is a review for Final exam, I do not get extra credits for completing it (the review), or anything like that.

My coursecompass.com account expired today, and I can't use it for studying

I have 60 problems on review, and I just did 30 (used 12 sheets of printer paper for scratching). On couple of them, my answer != answer in the answer key. Some explanations and help will be appreciated :thumbsup:

#7)

3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:


Re-arrange to write it as: sqrt(x+3) = x-3
Square both sides: x+3 = x^2 - 6x + 9
Simplify and solve. You don't need the quadratic formula. You can factor this one. Bring everything to one side so that you have 0 on the other side.




#9)

(x+28)/(x+1) < 4
I can understand x != -1, but how did they get x !=8? What were the steps?


You need to solve the inequality for x. Multiply both sides by (x+1) and go from there.




#19)
Find slope and y-intercept.
x-5y - 5 = 0
This is probably so simple it's humiliating. What are the STEPS to finding slope and y-int? I do not need the answer, it's already given in answer key (slope = 1/5 and y-int (0,-1)).


Isolate for y: -5y = 5-x --> y = x/5 -1

You can also recognize that if the equation of a line is given in standard form (Ax + By = C), the slope is given by -A/B

The y intercept is found by setting x = 0.



#20)
Write the equation of the line in standard form.
x-int is -5
y-int is 3
I know std form is Ax+By = C, but I do not understand how to go about the steps? Answer is (3x-5y = -15).


Find the slope of the line (since you are given two points) and substitute it into the equation of the line. Substitute one of the two points to find your constant C.

See the problem above for information about the slope.


#29)
The base and height of a triangle must have a sum of 40 inches. Find the base and height of the triangle whose area is as large as possible.

This is probably also dead simple, but I am stuck. I know that
b+h = 40
Area = 0.5 * bh

Answer is base 20 in, height 20 in.


This is actually a calculus problem. Not sure why they would ask you this.


Here's the PDF with review:
Text

Thanks for your help, o allmighty math gurus! And may your respect++

 

[DrPizza]

#29 calculus isn't necessary to solve it.

you have b + h = 40. Rearrange and solve for one of the variables.

Then you have Area = 1/2 b*h
Substitute for one of the variables.
The area function is a parabola (a sad parabola)
Find the value of x on the axis of symmetry.

 

[frostedflakes]

Yup, you should end up with something like A=-0.5h^2+20h. Of course you can also have area in terms of b, just depends on what you substituted in; both will lead to the same answer. You can then use a graphing calculator to find the the peak of the parabola, and the x (h in this case) value that corresponds to that peak will maximize area. Then plug this h back into the b+h=40 to solve for b.

Best of luck with finals man.

 

[MegaVovaN]

Thanks guys! I did all the problems with your help.
As for #29, my calculator (ti-83) graphed some weird, almost vertical line, so I found vertex manually (asked for help in class).

I am now doing next 30 problems, hopefully I won't have anymore questions.

respect++

 

[RapidSnail]

Basic questions:

1) Is there something I'm missing about the order of operations for question 7? Why can't the root be found by:

9 + x + 3 = x^2

Are you supposed to isolate the radical first before eliminating it?

2) In the standard form of a linear equation, A and B must be integers?

3) For problem 29 could you not just find which combination of naturals equaling 40 has the greatest product (since 1/2 = k)?

1+39 - 1*39
2+38 - 2*38

...

19+21 - 19*21
20+20 - 20*20

 

[aplefka]

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

 

[Taejin]

Originally posted by: MegaVovaN
Hi all!
First of all, this is not homework, it is a review for Final exam, I do not get extra credits for completing it (the review), or anything like that.

My coursecompass.com account expired today, and I can't use it for studying

I have 60 problems on review, and I just did 30 (used 12 sheets of printer paper for scratching). On couple of them, my answer != answer in the answer key. Some explanations and help will be appreciated :thumbsup:

#7)

3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

Alternatively you can make sqrt(x+3) = y.
Then you have 3 + 3 + sqrt(x+3) = x + 3.
Which is then 6 + y = y^2.
y^2 - y - 6 = 0
(y - 3) (y + 2) = 0
y = 3, y = -2
x = 3^2 - 3, x = (-2)^2 - 3
x = 6, or x = 1.
x = 1 doesn't work in the original equation, so that answer is removed.
x = 6 checks out.



#29)
The base and height of a triangle must have a sum of 40 inches. Find the base and height of the triangle whose area is as large as possible.

This is probably also dead simple, but I am stuck. I know that
b+h = 40
Area = 0.5 * bh

Answer is base 20 in, height 20 in.

Well, I don't know if there is a formal algebraic proof for this problem, but this problem strikes me as common sense more than anything else.

A similar problem is, you have a rectangle who's combined sides = 40. What's the biggest area?
Answer, a square.

If you assume your base is infinitely close to 0, and your height is infinitely close to 40, then you have a total area of 0. Same with if your base is infinitely close to 40, and your height is infinitely close to 0.

Now, try 1 and 39
1*39 = 39
2*38 = 76
3*37 = 111
4*36 = 144
.
.
.
10*30 = 300
.
.
.
.
15*25 = 375
.
.
.
19*21 = 389
20*20 = 400
21*19 = 389

To me its pretty obvious that the values 'merge' at a center point where the area becomes the largest possible.




Here's the PDF with review:
Text

Thanks for your help, o allmighty math gurus! And may your respect++

 

[Taejin]

Originally posted by: aplefka

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

yes. 1 doesn't work because of the sqrt's effect on the equation. If you plug 1 in, then you get 4 = 1

 

[Saint Michael]

Originally posted by: Taejin

Originally posted by: aplefka

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

yes. 1 doesn't work because of the sqrt's effect on the equation. If you plug 1 in, then you get 4 = 1

5=1.

This is one of the types of equations that makes me dislike math. One of the refreshing things about math ususally is that you can get nice clean answers by following basic rules. In situations like this (and especially with inequalities) much tinkering is required before a clean answer can be arrived at.

 

[Kyteland]

Originally posted by: Saint Michael

Originally posted by: Taejin

Originally posted by: aplefka

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

yes. 1 doesn't work because of the sqrt's effect on the equation. If you plug 1 in, then you get 4 = 1

5=1.

This is one of the types of equations that makes me dislike math. One of the refreshing things about math ususally is that you can get nice clean answers by following basic rules. In situations like this (and especially with inequalities) much tinkering is required before a clean answer can be arrived at.

3 + sqrt(x+3) = x and x^2-7x+6 = 0 are not the same equations although you derive one from the other. Although they are similar, they do not have the same domains. That is why you have to plug the potential answers from the second equation back in to the first one. 6 is the only correct answer to this question.

 

[aplefka]

Originally posted by: Taejin

Originally posted by: aplefka

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

yes. 1 doesn't work because of the sqrt's effect on the equation. If you plug 1 in, then you get 4 = 1

Dammit. I've always been too lazy to do that unless instructions remind me. I worked for the potential answers, can't you do the easy part yourself since you are the teacher? :|

 

[MegaVovaN]

Originally posted by: aplefka

Originally posted by: Taejin

Originally posted by: aplefka

Originally posted by: MegaVovaN
#7)
3 + sqrt(x+3) = x
Find x.
I calculated that x = -3 and x = -4, but they both do not check. Answer key says x = 6, and 6 checks, but how to get 6?
Solved it again, using quadratic formula, and answers are -2 and -7/2 :roll:

I don't know if your key is right because it should be 6 and 1. You can't just get one answer with that. I end up with 6 and 1.

Step 1: 3 + sqrt(x + 3) = x
2: sqrt(x + 3) = x - 3
3: sqrt(x + 3)^2 = (x - 3)^2
4: x + 3 = x^2 - 6x + 9
5: 0 = x^2 - 7x + 6
6: 0 = (x - 6)(x - 1)
Solution: 6, 1

Maybe I'm wrong?

yes. 1 doesn't work because of the sqrt's effect on the equation. If you plug 1 in, then you get 4 = 1

Dammit. I've always been too lazy to do that unless instructions remind me. I worked for the potential answers, can't you do the easy part yourself since you are the teacher? :|

Do you mean plugging the answers back in the original equation is the hard part? :laugh:
By the way, answer key is correct. Only 6 checks out in original formula. For example of this, see problem #3 in PDF. As you solve it, you'll get 2 answers, but only 1 is within constraints of the problem. Here is #3:

A ladder is 1 foot longer than its vertical height along the side of the house. The distance of the base of the ladder from the house is 7 feet less than its vertical height from the ground. How far up the house does the ladder reach?

I drew a triangle as to what sides equal. Triangle
Use Pythagorean theorem, and you'l lget quadratic equation. Factor it, and only 1 of answers will check within problem constraints (you can't have a negative side length)

Originally posted by: RapidSnail
Basic questions:

1) Is there something I'm missing about the order of operations for question 7? Why can't the root be found by:

9 + x + 3 = x^2

Are you supposed to isolate the radical first before eliminating it?

2) In the standard form of a linear equation, A and B must be integers?

3) For problem 29 could you not just find which combination of naturals equaling 40 has the greatest product (since 1/2 = k)?

1+39 - 1*39
2+38 - 2*38

...

19+21 - 19*21
20+20 - 20*20

1. I messed up order of operations before, thus it did not work (no correct answer). There are probably several ways to solve #7, but if you isolate square root part and square both sides, it's very easy to find x.

2. AFAIK, not necessarily integers. You can have fractions.

3. Yes I could, however we are studying quadratic equations and this problem can be solved with quadratics. Here's how we solved #29 in class:


b+h = 40
Area = y = 1/2 bh

deriving a 3rd equation: b = 40 - h

This is going to be a quadratic equation with U shape, up or down (it is down but I don't have flipped U shape on keyboard).
What we are looking for here is MAX or MIN of parabola. In other words, vertex.

substitute 3rd equation into first eq:
y = 1/2 (40-h)(h)
y = 1/2 h(40-h)
y = 20h - 1/2 h^2

A = -1/2 h^2 + 20h <-- quadratic equation. It is U shaped and opens down, like sad face (thanks DrPizza for analogy) because A is negative (-1/2)

in this equation, a = -1/2
b = 20
c = 0

vertex is h = (-b)/(2a)
h = (-20)/(2* (-1/2))
h = -20/-1
h = 20

One of the sides is 20. Going back to one of my original formulas, b+h = 40
b + 20 = 40
b = 20

So sides are 20 and 20. Taejin is correct that biggest area is a square.
This quadratic formula way to solve it is useful if we are given some weird numbers, like "area is 1.4536 units"

 

[RapidSnail]

Thanks MegaVovaN.

:thumbsup: