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## #1 Re: Help Me ! » Solving Radical Equations - I » 2015-04-01 04:03:48

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.'

## #2 Re: Help Me ! » Solving Radical Equations - I » 2015-04-01 03:31:06

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

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.

## #3 Help Me ! » Solving Radical Equations - I » 2015-03-31 06:02:30

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

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

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

## #4 Re: Help Me ! » Simplifying » 2015-02-06 07:47:14

Awesome! Thanks everyone!

## #5 Re: Help Me ! » Simplifying » 2015-02-06 03:20:50

(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?

## #6 Re: Help Me ! » Simplifying » 2015-02-01 04:49:06

{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?

## #7 Re: Help Me ! » Simplifying » 2015-01-31 04:43:22

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.'

## #8 Re: Help Me ! » Simplifying » 2015-01-26 05:34:41

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

## #9 Help Me ! » Simplifying » 2015-01-22 02:38:57

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)]

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

## #10 Re: Help Me ! » Radicals and Roots » 2014-10-07 02:57:02

#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?

## #11 Re: Help Me ! » Radicals and Roots » 2014-10-07 02:55:55

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?

## #12 Re: Help Me ! » Radicals and Roots » 2014-09-24 10:56:51

#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)

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

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.

## #13 Re: Help Me ! » Radicals and Roots » 2014-09-24 03:50:07

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)]

7. SQRT(5) * SQRT(5) * SQRT(9)
A 27
B 15
C SQRT(15)
D 25
E 5[SQRT(9)]
F SQRT(225)

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)]

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.

## #14 Re: Help Me ! » Radicals and Roots » 2014-09-23 09:08:12

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.

## #15 Re: Help Me ! » Radicals and Roots » 2014-09-23 05:22:48

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.

## #16 Re: Help Me ! » Radicals and Roots » 2014-09-23 02:39:52

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?

## #17 Re: Help Me ! » Radicals and Roots » 2014-09-22 14:04:19

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?

## #18 Help Me ! » Radicals and Roots » 2014-09-22 08:45:08

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)

## #19 Re: Help Me ! » Quick Question » 2014-09-19 02:09:53

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

## #20 Re: Help Me ! » Quick Question » 2014-09-18 14:48:09

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

## #21 Help Me ! » Quick Question » 2014-09-15 10:45:56

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.

## #22 Re: Help Me ! » Factoring Polynomials of Higher Degree » 2014-07-27 05:20:01

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)

## #23 Re: Help Me ! » Factoring Polynomials of Higher Degree » 2014-07-26 07:24:02

Thanks ShivamS, but I want to understand how to find the answer. Like what steps am I supposed to take.

## #24 Re: Help Me ! » Factoring Polynomials of Higher Degree » 2014-07-26 03:36:03

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

## #25 Help Me ! » Factoring Polynomials of Higher Degree » 2014-07-23 07:18:30

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.