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Q2.

Divide both sides by 2:

How should I carry on from here?

Q3.

Well, I read my lesson and it says "no positive square root can ever possibly equal a negative number." But if we could go on with it I would say the answer is 'B.'

So sorry about the misunderstanding! Q4 is indeed 2x^2. I see the latex page and I'll be going with that from now on.

2.

A 1

B 9/4

C -1/2

D 17/2

E -9/4

F none of the above

3.

A 0

B 4

C-4

D-2

E no solution

F all real numbers

My Answer: E

4.

A

B {-1,1}

C {-1/2,1}

D 1

E 0

F none of the above

5.

A no solution

B {-2,-3}

C {2,3}

D -2

E 3

F 0

NOTE:

Currently editing this post to make sure that the latex comes out right.

**demha**- Replies: 7

Hey guys I'm back! I need some help understanding how to solve some equations and checking the ones I solved already to see if they are right. For the ones I need help solving, I would prefer to be helped understand how to get the answer rather than just being given the answer. As usual I only have about 20 math problems so I'l be posting 5 of them at a time. Here is the first part:

1. [SQRT(x - 3)] - 7=0

A -46

B 10

C 52

D 4

E {4,10}

F none of the above

My Answer: C

2. 2[SQRT(w + 4)]=5

A 1

B 9/4

C -1/2

D 17/2

E -9/4

F none of the above

not sure how to solve

3. [SQRT(x)]=-2

A 0

B 4

C-4

D-2

E no solution

F all real numbers

My Answer: E

4. [SQRT(2x2 - 1)]=x

A[SQRT(1/3)]

B {-1,1}

C {-1/2,1}

D 1

E 0

F none of the above

not sure how to solve

5. [SQRT(2x^2 + 5x + 6)]=x

A no solution

B {-2,-3}

C {2,3}

D -2

E 3

F 0

not sure how to solve

Awesome! Thanks everyone!

(sqrt 3).3(sqrt 3) - 15 (sqrt 3) + 7 (sqrt 3) - 35

3 x 3 - 15 (sqrt 3) + 7 (sqrt 3) - 35

9 - 8(sqrt 3) - 35

-8(sqrt 3) - 26

So the answer will be D?

{3[SQRT(3)] + 7} [SQRT(3) - 5]

First Part:

(sqrt 3).3(sqrt 3) + 15 (sqrt 3) + 7 (sqrt 3) + 35

Second Part:

3 x 3 + 15 (sqrt 3) + 7 (sqrt 3) + 35

Third Part:

9 + 22(sqrt 3) + 35

Fourth Part:

22(sqrt 3) + 44

Is this right?

For #1 I still don't understand how to do it. Do you think you could show an example with a similar problem?

For #5 I am getting the answer 'F.'

For #3 I came up to {4x[SQRT(7)]}/14y which is answer A. Is that the correct answer?

**demha**- Replies: 11

Three of these I am not sure how to do, one of them I did my self and posted here just to make sure I got the right answer.

1. {3[SQRT(3)] + 7}[SQRT(3) - 5]

A -35

B -32

C 3[SQRT(3)] + 2

D -8[SQRT(3)] - 26

E 3[SQRT(3)] - 12

F 3[SQRT(9)] - 8[SQRT(3)] - 35

2. 4/[SQRT(5)]

A 4/[SQRT(5)]

B{4[SQRT(5)]}/25

C 4/25

D [SQRT(5)]/25

E {4[SQRT(5)]}/5

F 100/[SQRT(5)]

Answer: E

3. 4x/{2y[SQRT(7)]}

A{4x[SQRT(7)]}/14y

B 28x/14y

C 14x/y

D 4x/14y

E [SQRT(14)}/14y

F 4x/y[SQRT(7)]

4. A weight hangs on a spring. The vibration period of that weight is the time in which it makes a complete cycle in motion. Suppose the relationship between the vibration period "T" (in seconds) and the weight "w" (in kilograms) is given by T = 2 pi SQRT(w / 200). Find the period T for a spring with a hanging weight of 2.0 kilograms.

A 0.3 seoconds

B 0.6 seconds

C 0.9 seconds

D 1.2 seconds

E 1.5 seconds

F 1.9 seconds

5. Suppose that the volume "V" (in cubic feet) of a blimp with radius "r" is given by V = 4/3 * pi * r^3. If the blimp has a volume of 10,000 cubic feet, find the approximate radius of the blimp.

A 13.98 feet

B 14.5 feet

C 62864.79 feet

D 5863.2 feet

E 29 feet

F 13.367

I am guessing of setting up the equation as:

10,000 = 4/3 x pi x r^3

#14

4x^2 - 16y^2

Square it to become:

2x - 4y

Which is answer 'A'?

#15

SQRT(8y) - SQRT(36)

2 x 4 9 x 4

2[SQRT(4y)] - 9[SQRT(4)]

Am I doig it right?

12.

I watched a video tutorial and got an idea of how to set it, tell me if I'm doing it right:

Then set it as:

Is this the correct way?

#9

Oh I get it! Instead of saying 1[SQRT(2)] it is just SQRT(2). So since the equation equals up to 0, there is nothing left, so the real answer is 'D'.

#10

This is how I thought of doing it:

2[SQRT(5)] + 3[SQRT(20)]

In the two sqrts, I looked for a same number that went into both of them, which was 5. So then I set it up like this:

2[SQRT(5)] + 3[SQRT(20)]

1 x 5 4 x 5

I take the 1 and 4 and square them to get 1 and 2, then add it together like:

1+2[SQRT(5)] + 2+3[SQRT(5)]

For the final answer of 8[SQRT950]

Is this a correct way of doing it and could it be applied to all similar problems?

---

Here are some more:

11. SQRT(12) - SQRT(27)

A SQRT(3)

B SQRT(12)

C SQRT(27)

D-SQRT(19)

E -SQRT(3)

F -SQRT(12)

My Answer: 'E'

12. SQRT(90a^4 b^7)

A 10a^2 b^3 [SQRT(9b)]

B 9a^4 b^6 [SQRT(9b)]

C 9a^2 b^3 [SQRT(9b)]

D 3a^2 b^3 [SQRT(10b)]

E 9a^4 b^6 [SQRT(10b)]

F 3a^4 b^6 [SQRT(10b)]

I'm not sure how to solve this one. It probably looks confusing. How can I set it up like you do to show a number that is to the power instead of adding this ^?

13. [SQRT(13) + SQRT(8)][SQRT(13) - SQRT(8)]

A 233

B 13 + SQRT(8) - SQRT(13)

C 21

D 169 + SQRT (64)

E 5

F 28

My Answer: 'E'

14. [SQRT(2x) + SQRT(4y)][SQRT(2x) - SQRT(4y)]

A 2x - 4y

B 2x + [SQRT(16xy)] + 4

C 2x - 8xy + 4y

D 2x - 2xy + 4y

E 2x - 2[SQRT(3xy)] + 4

F 2xy + 4xy

I want to show you how I got and how I solve it:

(2x + 4y)(2x - 4y)

2x * (2x + 4y) 4y * (2x - 4y)

(4x^2 + 8xy)(8xy - 16y^)

Cancel out the 8xy and left with:

4x^2 - 16y^2

Normally I would square this then add/subtract them to get a final answer. Since I have the ^2, what do I do instead? I don't see what my final answer would be in the list.

15. SQRT(2)[SQRT(4y) - SQRT(18)]

A 2y - 3

B SQRT(8y) - 6[SQRT(2)]

C SQRT(8y) - 6

D 2[SQRT(2y) - 6]

E SQRT(2y) - 3[SQRT(2)]

F 2[SQRT(2y)] - 6

Not sure how to solve this one. I thought of doing it this way:

SQRT(2)[SQRT(4y) - SQRT(18)]

work on [SQRT(4y) - SQRT(18)] first:

SQRT(4y) - SQRT(18)

2 x 2 = 4 9 x 2 = 18 so

2[SQRT(2y)] - 9[SQRT(2)]

so now I am left with:

SQRT(2){2[SQRT(2y)] - 9[SQRT(2)]}

This is where I don't know how to proceed. Normally I would square the 9 and 2 then add/subtract them. But I'm just confused at this point.

So for #4 the answer would be 'D' 25a^2

And as for #5, I would then assume the answer is 'A' -27. But how is it in an earlier one (-125^2/3) the answer did not have a minus, the minus was turned into a positive?

Here are more:

6. 4[SQRT(3)] - 7[SQRT(3)]

A -9

B-11[SQRT(3)]

C-3[SQRT(3)]

D 9

E 11[SQRT(3)]

F 2[SQRT(3)]

Answer: 'C'

7. SQRT(5) * SQRT(5) * SQRT(9)

A 27

B 15

C SQRT(15)

D 25

E 5[SQRT(9)]

F SQRT(225)

Answer: 'B'

8. SQRT(196) + SQRT (441)

A 35

B 6[SQRT(49)]

C 7[SQRT(13)]

D 25

E 25[SQRT(12)]

F 5[SQRT(12)]

Answer: 'A'

9. SQRT(2) - 3[SQRT(2)] + 2[SQRT(2)]

A -4[SQRT(2)]

B SQRT(2)

C -SQRT(2)

D 0

E 4[SQRT(2)]

F -2[SQRT(2)]

Aswer: 'B'

10. 2[SQRT(5)] + 3[SQRT(20)]

A 8[SQRT(20)]

B 11[SQRT(20)]

C 5[SQRT(5)]

D 8[SQRT(5)]

E 14[SQRT(5)]

F 5[SQRT(20)]

I wasn't sure on how to do this one. I thought of a possible answer of 'D'. Weather this is correct or not, I would like to know how to solve this.

So that will be answer x^(3/2)

So you pretty much take the fraction and reduce it, am I correct? (From 6/4 to 3/2)

Here are some more:

4. (-3125a^5)^(2/5)

A 15a^3

B 5a

C 10a^2

D 25a^2

E 75a^3

F 125a^3

This one I was't really sure how to do since the larger number in parenthesis has the 5th power.

5. -2187(3/7)

A -27

B 312

C 21

D 27

E -21

F -312

My answer for this s 'D' 27.

So from what I understand, this how I THINK it goes:

-125^2/3

it becomes:

15,625 and I'm left with 1/3. So I need to find a number timed by itself 3 times to get 15,625 which is the answer, therefore being 'B' 25?

SO for that, I am given answers:

A 5

B 25

C -5

D -25

E 253

F 75

For the next problem, I am given:

x^(6/4)

with possible answers of:

A 6/4x

B x(2/3)

C 6x/4

D x(2)

E x(4/6)

F x(3/2)

How would I solve this with no number for 'x'?

EDIT:

For the first part, I believe it would be -25 not 25, so the answer would be 'D' -25.

So for that, my lesson gives me these answers to choose from:

A 3.83

B 4.5

C 18

D 10.90

E 4.8

F 1080

After square rooting twice, I got 3.8336 and a long list of numbers. So I just took the first three digits and matched it with 'A' as my answer.

Now another one goes like this:

(-125)^2/3

So, I believe it would be (-125)^2 which makes it 15,625 making it a positive. So it will be set up like:

x^3 = 15,625

How should I square root this?

Hi bobbym,

Where would I move on from there?

Should I do:

(6 * 6 * 6) = (216)^1/4? If so, where do I go from here?

**demha**- Replies: 16

Need some more help. This is the first part. I would like an explanation of how to solve it:

Simplify

6^(3/4)

3.9t - 15.6 = 0

SO we are solving for 't' here. Reordering:

-15.6 + 3.9t = 0

Add 15.6 to both sides:

15.6 + (-15.6) + 3.9t = 0 + 15.6

3.9t = 15.6

Divide each side by 3.9 for a final answer of:

t = 4

Answer: 'D'

So with that I'll just solve for t and get my answer,correct?

**demha**- Replies: 5

I was given this word problem:

Suppose a football is kicked from the ground and its height in feet above the ground is given by h = -3.9t^2 + 15.6t. The height above the ground is "h". The time in seconds, after the ball is kicked, is "t". How long is the ball in the air?

A 10 seconds

B 5 seconds

C 18 seconds

D 4 seconds

E 15.6 seconds

F 3.9 seconds

I took

-3.9t^2 + 15.6t

and then factored -1t to be left with:

-1t(3.9t + 15.6)

From here I'm not all too sure where to go.

It's still not making too much sense for me. I tried putting in to this:

(a + b)(a2 – ab + b2)

But it just comes up into a huge mess:

(x^3)^3 + 1^3

((x^3)^3 + 1^3)(((x^3)^3)2 – ((x^3)^3)(1^3) + (1^3)^2)

I come up with (x^3)^3 + 1^3 but I'm not entirely sure where to go from here.

**demha**- Replies: 9

Hey guys, I need help with Factoring Polynomials of Higher Degree. I got around 20 questions and I'll post 5 at a time. For the first five I answered all but one:

1. 125x^3 + 64

A (5x - 4)(25x^2 + 20x - 16)

B(5x + 16)(25x^2 - 20x + 2)

C(5x + 8)(25x^2 - 40x + 16)

D(5x + 4)(25x^2 - 20x + 16)

E (5x + 2)(25x^2 - 40x + 16)

F cannot be factored

My answer is D.

2. 216a^3 + 1

A (3a + 1)(24a^2 - 8a + 1)

B(6a + 1)(36a^2 - 6a + 1)

C(18a + 1)(2a^2 - 6a + 1)

D(4a + 1)(20a^2 - 4a + 1)

E (6a - 1)(36a^2 + 6a + 1)

F cannot be factored

My answer is B.

3. 7x^5 - 64y

A 7xy(x^2 - 8)(x^2 + 8)

B 7x(x^2 - 8)(x^2 + 7)

C 7xy(x^4 - 8)

D 7x(x^2 - 9)(x^2 + 1)

E 7(x5 - 9x)

F cannot be factored

My answer is F.

4. 27x^3 - 8

A (3x - 2)(3x + 2)2

B(3x - 2)(9x^2 - 6x + 4)

C(3x + 2)(3x^2 - 3x - 4)

D(27x + 2)(x^2 - 2x + 4)

E (3x - 2)(9x^2 + 6x + 4)

F cannot be factored

My answer is E.

5. x^9 + 1

A (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1)

B(x^3 + 1)(x^6 + 1)

C(x + 1)(x^2 - x + 1)(x^3-1)2

D(x + 1)3(x^2 - x + 1)3

E (x^3 + 1)(x^6 - x^3 + 1)

F cannot be factored

I am having a little trouble on this one. I'm not really sure how I would even begin with it.