Question

helppp

Answer 1

\[3\sqrt[4]{128}\]

Answer 2

no i have no idea how to do these

Answer 3

Ok, the thing you need to know is that that little tiny 4 there is called the index of the radical. In essence, what you have up there is this\[3(128)^{\frac{ 1 }{ 4 }}\]Did you know that? Are you familiar with the exponential ways to write radical expressions?

Answer 4

no but now i am

Answer 5

Do you have choices from which to pick for solutions or no?

Answer 6

Any time you have a radical with a little number stuck in the dip of it, that's called the index of the radical. It is rewritten to exponential form like I showed you. Here's another couple of examples so you get the idea.\[\sqrt[5]{32}=32^{\frac{ 1 }{ 5 }}\]\[\sqrt[4]{2^{3}}=2^{\frac{ 3 }{ 4 }}\]So you can kind of see how this goes.

Answer 7

What you have, in words, is three times the fourth root of 128. You have to find the 4th root of 128 on your calculator. Do you know how to do that?

Answer 8

Do you have to have an actual answer or are you just rewriting these in exponential form?

Answer 9

yeah i see.. i dont know how to do that

Answer 10

Do you have to have an actual answer for this or are you just converting it to exponential form?

Answer 11

all the answers are in radical form

Answer 12

Give me examples of the answers they gave you for this one, please.

Answer 13

\[2\sqrt[4]{8}\]
\[5\sqrt[4]{8}\]

Answer 14

Ok, let me examine this for a minute or two.

Answer 15

ok

Answer 16

Ok, what they are asking you to do here is to simplify and leave your answer in radical form. In essence, what this problem wants you to do is find 2 numbers that, multiplied together equal 128, but one of them has to have a perfect fourth root in it. This takes poking at your calculator to find. I found that 128 = 16 * 8. 16 has a perfect fourth root of 2 in it (2^4 = 16). So rewrite your radical to reflect that\[3\sqrt[4]{16*8}=3\sqrt[4]{(2)^{4}*8}\]

Answer 17

The 2^4 comes out of the radical as a 2 and you're left with an 8 under the radical sign still, right? But you already had a 3 out there so now you have\[3*2\sqrt[4]{8}\]Multiply the 3 and the 2 to get 6 so your final answer should be\[6\sqrt[4]{8}\]Is that one of your choices, I hope?

Answer 18

it is! thank you so much for your help

Answer 19

You're welcome! I will gladly help with more so you can grasp the concept a little better.

Answer 20

should i post another one on here?

Answer 21

can you solve that on a calculator

Answer 22

You can solve on a calculator the number value, but it will not leave your answer in radical form. What you CAN do on your calculator is find the numbers you need to multiply together and check them for a 4th root or a cubed root, etc. Yes, post here with more, if you have them!

Answer 23

\[3\sqrt[4]{3}-8\sqrt[3]{3}+5\sqrt[4]{3}+4\sqrt[3]{3}\]

Answer 24

Oh boy! ; ) What do the directions tell you to do with this?

Answer 25

simplify

Answer 26

Ok, let me get a couple done and then explain them one by one, ok?

Answer 27

ok

Answer 28

All the radicands are the same (the 3 under the radical sign), so that's a good start. The rule is that you can only add or subtract like radicands that also have the same index. So that's where we need to start.

Answer 29

Let's rearrange this so it is easier to see the "like" stuff

Answer 30

\[3\sqrt[4]{3}+5\sqrt[4]{3}-8\sqrt[3]{3}+4\sqrt[3]{3}\]We have two terms that have a fourth root and two that have a cub root. As long as the radicands are the same, we can add them:\[3\sqrt[4]{3}+5\sqrt[4]{3}=8\sqrt[4]{3}\]That's the first part. You try to simplify the next two terms. They are also "like" so you can combine them. What do you get?

Answer 31

\[-4\sqrt[3]{3}\]

Answer 32

Yep! So putting them together gives you your solution:\[8\sqrt[4]{3}-4\sqrt[3]{3}\]And that's your result! See how to do that?

Answer 33

oh yeah!\[\sqrt{128}+2\sqrt{112}-2\sqrt{98}-6\sqrt{28}\]

Answer 34

All these have the same index as well. It is 2 because those are square roots. What you need to do in a situation like this is to make the radicands all the same. That means you need to simplify the 128, 112, 98, and 28 and see what they are in simplest form. Let's start with the 128. We need to find two numbers, one of them a perfect square, that multiply together to get 128. Let's see. The multiples of 128 are 2 * 64, 4 * 32, 8 * 16. Looking at those options, which choice has one of the numbers as a perfect square in it? Only one perfect square; it will make the simplification process easier.

Answer 35

8?

Answer 36

Look at 2 * 64. 2 does not have a perfect square in it but 64 does. It's 8. Is that what you meant?

Answer 37

yeah

Answer 38

Ok good then. What is that term then in simplest form?

Answer 39

\[8\sqrt{2}\]

Answer 40

Perfect. Let's move on to the next one, then.

Answer 41

\[2\sqrt{112}\]The options for your multiples are 2 * 56, 4 * 28, and 16 * 7. Which one of those has one perfect square in it?

Answer 42

16

Answer 43

That's right. Now simplify that expression, and don't forget the 2 that is already out front. What does that simplify down to?

Answer 44

\[8\sqrt{7}\]

Answer 45

Perfect!! Are you sure you need help?! ; ) What you have so far is\[8\sqrt{2}+8\sqrt{7}\]Now move on to the next one. The 98 one. The only choices for 98 are 2 * 49. Is one of those a perfect square?

Answer 46

49

Answer 47

Ok, good...so simplify that expression all the way, remembering the 2 out front that's already there. What do you get for that one?

Answer 48

\[14\sqrt{2}\]

Answer 49

You're good! Now what's the next one? Can you give me that in its simplest form on your own?

Answer 50

\[12\sqrt{7}\]

Answer 51

Yes! So what you have altogether is\[8\sqrt{2}+8\sqrt{7}+14\sqrt{2}-12\sqrt{7}\]Remember that you can add and subtract like radicands. So what do you get when you do that?

Answer 52

\[22\sqrt{2}-4\sqrt{7}\]

Answer 53

You got it!

Answer 54

See? That wasn't bad at all!

Answer 55

it dosnt match any of the answers

Answer 56

\[-6\sqrt{2}-4\sqrt{7}\]

Answer 57

Then you must have one of your signs off in the original problem or something, because that is definitely the answer to the problem you gave. There is no other way to answer it.

Answer 58

the original equation is right

Answer 59

Oh...oops it was my fault, I changed one of the signs. My bad. Here's what you have (in the correct form with the correct signs!)

Answer 60

its all good

Answer 61

\[8\sqrt{2}+8\sqrt{7}-14\sqrt{2}-12\sqrt{7}\]I had that 14 with a positive sign, not a negative one. It works out that way, now. Try it.

Answer 62

\[-6\sqrt{2}-4\sqrt{7}\]

Answer 63

thats it

Answer 64

That's right. Again...see, not too bad, right?

Answer 65

yep.. im going to work on all the easy questions ill be back

Answer 66

Cool