Radicals and Roots

demha · 2014-09-22 08:45:08

Need some more help. This is the first part. I would like an explanation of how to solve it:
Simplify
6^(3/4)

Last edited by demha (2014-09-22 08:46:16)

bobbym · 2014-09-22 09:00:42

Hi;

demha · 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?

Bob · 2014-09-22 19:03:46

hi demha,

That's right. Then you want x so that

There's no integer solution to that so you'll have to use a calculator.

Either square root twice or use the y^x button ( 216 y^x 4 = ) Some makes of calculator (incorrectly) label this button x^y.

You could also find the fourth root of 6, and then cube it.

Bob

demha · 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?

Bob · 2014-09-23 04:04:55

Answer A looks right to me.

You can square and then cube root, but also cube root first and then square. This makes the problem easier because it keeps the numbers lower.

Then square it.

If you really want to cube root 15625, then reduce it to prime factors:

15625 = 5 x 3125 = 5 x 5 x 625 = 5 x 5 x 5 x 125 = 5 x 5 x 5 x 5 x 25 = 5 x 5 x 5 x 5 x 5 x 5 = (5x5) x (5x5) x (5x5)

Bob

demha · 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.

Last edited by demha (2014-09-23 05:27:43)

Bob · 2014-09-23 08:06:36

hi demha,

B 25 is right. You want 25 x 25 x 25 = +15625

NOT -25 x - 25 x -25 = -15625

As this next has to be true for all x, let's explore a particular value of x. I want to be able to find a 4th root so I'll choose x = 81 (as 4th root is 3)

Now write that as 24 3s multiplied together as 6 blocks of 4

Now re-write as 4 blocks of 6

and find the 4th root

So which answer is the same ?

Write it as 2 blocks of 6 that is square rooted

and then as 3 blocks of 4 that is square rooted

What this shows is that you can simplify the exponent as a fraction

What is true for 81 must still be true when it is replaced by x.

edit: OK, I suppose you want me to prove it.

Bob

demha · 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.

Last edited by demha (2014-09-23 09:08:43)

Bob · 2014-09-23 18:27:59

Q4. ^5 means you have to raise to the 5th power and ^(1/5) means you have to take the 5th root; so these two cancel out just like they would with ordinary fractions.

Q5. There's a minus sign to account for too so D won't be right.

Bob

demha · 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)]

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.

Bob · 2014-09-24 08:31:12

Q4. D is good.

Minuses

Don't forget that -5 x -5 = + 25 but -5 x -5 x -5 = - 125. In Q5 the powers are all odd so the minus is there throughout.

In the -125^2 question, squaring gets rid of the minus so everything is + after that.

Q6. Correct.

Q7. Correct.

Q8. Correct.

Q9. I don't agree with B. Suppose you put x instead of root 2

x - 3x + 2x

Q10. You can re-write root(20) as root(4x5) = root(4) x root(5) = 2 root(5). So answer D is right!

Bob

demha · 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)

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.

Bob · 2014-09-24 16:56:51

#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'.

Correct

#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?

Correct

---
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'

Correct

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 ^?

Use Latex. There's a tutorial on the help page. Bit pressed for time now. I'll go through it in more detail next week.

Split into bits that will easily square root and bits that won't.

= root (9 x 10 x a^4 x b^6 x b) = root(9 x a^4 x b^6) x root(10b) = ...

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'

Correct

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.

Right idea to multiply and cancel the 'middle terms' but you've forgotten the square root. So root each of these expressions and it will work.

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.

Multiply the root 2 into the bracket:

root(2) x root(4y) - root(2) x root(18) = root(8y) - root(36) = .....

Bob

demha · 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?

demha · 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?

Bob · 2014-10-07 03:26:31

hi demha

Q12. Try to split into pairs that will square root eg. 9a^2 = (3a)(3a)

So we have

Now we have pairs that will square root:

btw: you can do a SQRT with

Q14 is correct.

Q15. SQRT(8y) - SQRT(36) = SQRT[4 x 2y] - 6 = 2 SQRT(2y) - 6

Bob

Math Is Fun Forum

#1 2014-09-22 08:45:08

Radicals and Roots

#2 2014-09-22 09:00:42

Re: Radicals and Roots

#3 2014-09-22 14:04:19

Re: Radicals and Roots

#4 2014-09-22 19:03:46

Re: Radicals and Roots

#5 2014-09-23 02:39:52

Re: Radicals and Roots

#6 2014-09-23 04:04:55

Re: Radicals and Roots

#7 2014-09-23 05:22:48

Re: Radicals and Roots

#8 2014-09-23 08:06:36

Re: Radicals and Roots

#9 2014-09-23 09:08:12

Re: Radicals and Roots

#10 2014-09-23 18:27:59

Re: Radicals and Roots

#11 2014-09-24 03:50:07

Re: Radicals and Roots

#12 2014-09-24 08:31:12

Re: Radicals and Roots

#13 2014-09-24 10:56:51

Re: Radicals and Roots

#14 2014-09-24 16:56:51

Re: Radicals and Roots

#15 2014-10-07 02:55:55

Re: Radicals and Roots

#16 2014-10-07 02:57:02

Re: Radicals and Roots

#17 2014-10-07 03:26:31

Re: Radicals and Roots

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