ok

- anonymous

ok

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

Simplify each term first. Add/Subtract with radicals works like variables. Only exact same radicals can be added together. You add numbers outside the radical.

- anonymous

You may not know yet, but sqrt of a variable simplifies like this:
sqrt(x^4) = x^2
sqrt(x^5) = x^2 sqrt (x)
sqrt(x^6) = x^3
Pattern?

- anonymous

Just take half of the power. That part is the power of the variable that goes outside. Any remainder stays inside.
\[\sqrt{x ^{5}}y ^{6}z ^{9} = x ^{2}y ^{3}z ^{4 } \sqrt{x}z\]

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## More answers

- anonymous

5/2 = 2 rem 1
6/2 = 3
9/2 = 4 rem 1

- Jhannybean

Break down each one using this method,then combine them all

- Jhannybean

\[\large \sqrt {75x^2y}-\sqrt{80xy^2}-\sqrt{48x^2y}+\sqrt{20xy^2}\]
let's start with \(\large \sqrt{75x^2y}\)
\[\large \sqrt{75x^2y} = \sqrt{5*5 * 3*x*x*y}\] for every pair of similar numbers, pull out only 1\[\large \sqrt{75x^2y} =\color{red}{ 5x \sqrt{3y}}\]

- Jhannybean

do you have a calculator with you?

- Jhannybean

if not,your computer always will have one :) just go to the start menu > search > type in calculator, you will need one handy in order to understand how to break down a number

- Jhannybean

So 75 = 25 * 3, yes? think of quarters, if you have 3 quarters, it equals 75 cents ,right? :)

- Jhannybean

now you have 25, and that = 5*5, right?

- Jhannybean

what do you mean? you have to break down a number into it's multiples TILL you get down to the very bare prime factors.
this is a little advanced but lets say you have the number 134, it's multipliers, or numbers that can be multiplied together to equal 134, are ONLY 1 and 134. so that means 134 is the bare minimum number

- Jhannybean

|dw:1370071609765:dw|

- Jhannybean

all the numbers circled are the numbers that cannot be broken down anymore.
All you do is gather up all these circled numbers, which are 5, 5 and 3., and multiply them, you will get 75 again :)

- Jhannybean

Does that make sense now?...

- Jhannybean

Yes you're right :) but you see how 40 can also be broken down to 20 x 2?

- Jhannybean

and then 20 can be broken down to 10 x 2 and 10 can be broken down to 5 x 2?
Take out a scratch piece of paper and try it with me!:) we'll work on this using a tree diagram :)

- Jhannybean

punch in 5 x 15 into your calculator, you'll see that it gives you 75, not 80 :(

- Jhannybean

oh! that can work too :)

- Jhannybean

and 1 can be broken down further....4x4..4s can be broken down to 2x2...getting it? :D

- Jhannybean

|dw:1370072135392:dw|

- Jhannybean

yeah! sometimes a big number can have various multipliers that will equal the same number :P you choose which multipliers you want to break down the numbers with

- Jhannybean

aslong as you can break them down to prime numbers,or numbers that can not be broken down further, i.e. 2's..3's..5's..7's.. prime numbers :P

- Jhannybean

can you try 48 for me? and tell me how you break it down? i'll cross check with you :)

- Jhannybean

good good,it can be broken down further :)

- Jhannybean

what are the factors of 12? 4? 6? 8?

- Jhannybean

nope. not the same number, just till it cannot be broken down anymore

- Jhannybean

you are correct :)

- Jhannybean

4 = 2* 2 :)

- Jhannybean

http://www.mathsisfun.com/prime-factorization.html this will help you understand prime factors :)

- Jhannybean

oh cause i had no more space lol... sorry about that :\ yes, 4 would have been broke down into 2*2 :) you were right,that's why i deleted it

- Jhannybean

So how you had originally factored it out was absolutely correct!

- Jhannybean

Shall we move onto the next one? :)

- Jhannybean

Good job!! you got this :D

- Jhannybean

yaaay!!!!~
Now we'll move on to "pairing" everything that's inside the square root. And you know what? It all works the same way! whether you have variables, or numbers like these :D

- Jhannybean

Yes :D it's easier than it looks now that you've understood how to break down the numbers :P (the hard part)

- Jhannybean

yess. \[\large \sqrt{75x^2y} = \sqrt{5*5*3*x*x*y}\]
You see how you have a pair of 5's, and x's inside the square root? you can pull them out as 1! so if you have 2 5's under a square root,you can pull out 1 of them, and since you have 2 x's under a square root,you can also pull out 1 of those :D then since everything else doesn't have another pair, you leave it inside the square root
So... \[\large \sqrt{75x^2y} = \color{red}{5x \sqrt{3y}}\]

- Jhannybean

\[\large \sqrt{\color{red}{2*2}*\color{blue}{2*2}*{x}*\color{green}{y*y}}\] you see how all the colored numbers and variables can be paired up and taken out of the square root? just like the ther one we did?

- Jhannybean

no no, remember you don't have a pair of x's,so you can't pull one out., although you are right about 2*2*y

- Jhannybean

ohh okay. Then you are right :)

- Jhannybean

\[\large \sqrt{80xy^2}=\color{red}{ 4y \sqrt{x}}\]

- Jhannybean

You got it! :D

- Jhannybean

Awesome!!! i'm so happy for you :) you're a fast learner!

- Jhannybean

\[\large \sqrt {75x^2y}-\sqrt{80xy^2}-\sqrt{48x^2y}+\sqrt{20xy^2}\] just so we have the problem here haha.

- Jhannybean

You can always rearrange the order ofthe numbers if it makes it easier for you :)

- Jhannybean

I'll solve it to :) we're working along side eachother! one minute :P

- Jhannybean

\[\large \sqrt{48x^2y} =\sqrt{\color{red}{2*2*2*2}*3*\color{blue}{x*x}*y}=\color{purple}{ 4x \sqrt{3y}}\]

- Jhannybean

you'reright!!!! you're good :D

- Jhannybean

\[\large \sqrt{20xy^2} = \sqrt{\color{red}{2*2}*5*x*\color{green}{y*y}}= \color{blue}{ 2y \sqrt{5x}}\]

- Jhannybean

mmhmm:) do the colors help btw?or are they distracting? D:

- Jhannybean

Now we're going to take ALL these things weve factored out, and combine them all. haha.

- Jhannybean

\[\large \sqrt {75x^2y}-\sqrt{80xy^2}-\sqrt{48x^2y}+\sqrt{20xy^2}\]\[\large \color{red}{5x \sqrt{3y}}-\color{red}{ 4y \sqrt{x}} - \color{purple}{ 4x \sqrt{3y}} +\color{blue}{ 2y \sqrt{5x}}\] yep you are right
Now we must combine things that aremultiplied by the same square root sign, so
\[\large (\color{red}{5x} \sqrt{3y} -\color{red}{4x} \sqrt{3y}) -4y \sqrt{x} +2y \sqrt{5x}\] you leave the square root signs alone and subtract the things highlighted in red

- Jhannybean

So when you subtract, you shall get \[\large x \sqrt{3y} -4y \sqrt{x} +2y \sqrt {5x}\]

- Jhannybean

we can't combine those because they have different factors under the square root...they will stay as is :)

- Jhannybean

Hypothetically,if it had been \[\large x \sqrt{3y}-4y \sqrt{3y}+2y \sqrt{3y}\] only THEN we could combine all the terms into one simplified form, but because theyre all different,we cannot combine them.
I don't even remember what the choices were lol

- Jhannybean

Yeah, it'd most likely be C

- Jhannybean

Do you feel like you learnt something here though? xD

- Jhannybean

Haha.3 hours? xD

- saifoo.khan

Wow, this is pretty long.

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