Question

Can someone help me understand how to simplify radicals? I have been through so many problems and I still don't understand. Thanks! Medal to best answer.

Answer 1

how r u doing so far

Answer 2

I have been doing practice problems but I have not gotten through one 100% correctly. I get lost in the steps to simplify! I can put a practice problem up if you want so you can help me with the process.

Answer 3

post an example, and i will go through the properties used to simplify

Answer 4

there really are only a couple properties to remember

Answer 5

\[\sqrt[3]{-128xy ^{4}}\]

Answer 6

remember any root sign also means the quantity underneath raised to the power of 1 over the order of the root... like this..

Answer 7

\[\sqrt[n]{a} = a^{1/n}\]

Answer 8

that is property 1

Answer 9

Okay, am I supposed to find the perfect square of -128?

Answer 10

\[\sqrt[3]{-128xy^4} = (-128xy^4)^{1/3}\]

Answer 11

oh I see, it has to do with the 3 on the \[\sqrt[3]{?}\]

Answer 12

the 1/3 that is

Answer 13

yeah that is the cubed root

Answer 14

same as raising the stuff to the (1/3) power

Answer 15

so, for instance if the number was a 2 like this:

\[\sqrt[2]{?}\]

it would be 1/2?

Answer 16

The square root is the same as raising to the (1/2) power

Answer 17

lol yep, but for 2, it is just square root, you dont write the 2

Answer 18

only numbers higher than 2

Answer 19

so, it wouldn't be 1/2?

Answer 20

no i am saying, when it is a 2, you just write
\[\sqrt{a} = \sqrt[2]{a}\]

usually dont put a 2 in there, only if the number is a higher root than 2 (the square root)

Answer 21

If you see this
\[\sqrt{x} = x^{1/2}\]

the 2 is implied

Answer 22

okay. so with the number 4 it would be 1/4?

Answer 23

yeah, the 4th root of something, is the same as something to the 1/4 power

Answer 24

okay, now back to the example...

Answer 25

\[\sqrt[3]{-128xy^4} = (-128xy^4)^{1/3}\]

Answer 26

The thing i usually do , if you want to simplify that, is leave the numbers as roots, and put the variables to the powers, like this....

Answer 27

\[\sqrt[3]{-128xy^4} =\sqrt[3]{-128}*(xy^4)^{1/3}\]

Answer 28

good on that part?

Answer 29

Yes

Answer 30

ok, now for the exponents

\[[a^b]^c = a^{b*c}\]
a power raised to a power, is the base to the powers multiplied together

Answer 31

that is property 2 , if you want to call it that

Answer 32

Great! So it's 4*1/3 ??

Answer 33

\[\sqrt[3]{-128xy^4} =\sqrt[3]{-128}*(xy^4)^{1/3} = \sqrt[3]{-128}*x^{1 * 1/3}*y^{4 * 1/3}\]

Answer 34

yep, move the power onto both the x and the y, like that above

Answer 35

4*1/3 = 1.33333

Answer 36

\[\sqrt[3]{-128xy^4} =\sqrt[3]{-128}*(xy^4)^{1/3} = \sqrt[3]{-128} * x^{1/3} * y ^{4/3}\]

Answer 37

nah, no decimal approximations, leave it as exact fractions

Answer 38

OOOH I get it. gotta keep em as fractions.

Answer 39

What's next boss?

Answer 40

yes, and that is all you can do to the variables, now for the cubed root of -128

Answer 41

\[\sqrt[3]{-128} = \sqrt[3]{-64 * 2}\]

Answer 42

see what i did there, 64 is a perfect cube 4*4*4

Answer 43

Okay. And I understand that you are supposed to get the number to it's most simple cubed root form. How do you know if it is at it's most simple? Also, how do you know a perfect square is 64?

Answer 44

a perfect cube is 64, 4 ^3 = 4 * 4 * 4

Answer 45

check if the number under the cubed root is a multiple of any of the perfect cubes, 2^3 = 8, or 3^3 = 27, or 4^3 = 64 ...and so on

Answer 46

and you notice 128 is 2 * 4^3

Answer 47

do you just divide by 2? or did you do 4^3. I run into the most issues when I am trying to find the perfect square.

Answer 48

yeah squares are easier, if it is an even number, just divide by 2 , then 2 again, then you will have 4 * (number) and square root of 4 is 2

Answer 49

okay that makes it easier.

Answer 50

just write your number out into its factors..like for 128

128 = 2 * 64 = 2 * 2 * 32 = 2 * 2 * 2 * 16

and there you see a perfect cube, 2^3 (3 2's)

Answer 51

okay

Answer 52

So we have: \[\sqrt[3]{-64*2}\]

Answer 53

\[\sqrt[3]{-128} = \sqrt[3]{(-4 * 4 * 4 * 2}) = \sqrt[3]{(-4)^3 * 2} = -4*\sqrt[3]{2}\]

Answer 54

you can pull out 7 2's from 128

128 = 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^7

Answer 55

so you get 2^6 * 2\[\sqrt[3]{128} = \sqrt[3]{2^6 * 2} = 2^{6/3}*\sqrt[3]{2} = 2^2\sqrt[3]{2} = 4 * \sqrt[3]{2}\]

Answer 56

\[\sqrt[3]{-128xy^4} = -4 *\sqrt[3]{2}*x^{1/3}*y^{4/3}\]

Answer 57

or you can just leave it as this if you dont want a root sign....

Answer 58

\[\sqrt[3]{-128xy^4} = -2^{7/3} *x^{1/3}*y^{4/3}\]

Answer 59

Okay. Isn't the x and the y pulled over to the left side though?

Answer 60

What do you mean? in front of the numbers?

Answer 61

multiplication can be done in any order

Answer 62

usually you write the constants first, then the variables

Answer 63

aren't the x and the y variables supposed to be pulled to the left side.

Answer 64

i'm not gettin what you mean

Answer 65

because this is an example I actually have the right answer to this and it is:
\[-4y \sqrt[3]{2xy}\]

Answer 66

i see what they want there ... just take out the perfect cubed variables and leave the rest underneath

Answer 67

\[\sqrt[3]{-128xy^4} = \sqrt[3]{(-2^6 * 2*x * y^3 * y} = -2^2*y* \sqrt[3]{2*x*y}\]

Answer 68

they just took out only the perfect cubes, instead of the whole thing like i did

Answer 69

alright. can we do one more but this time I'll try to do it to see if I can go through the process?

Answer 70

sure, can you post a new thread, and tag me, and immediately close it out so people dont swarm to it

Answer 71

yup