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\[2\sqrt{44x ^{3}}-\sqrt{7}-\sqrt{99x ^{3}}+\sqrt{63}\]

Combine like terms, right?
I don't see how to combine \[2\sqrt{44x ^{3}}\] and \[-\sqrt{99x ^{3}}\]

before combining see if you can simplify each term first

How so...?

Oh! like 11 goes into both 44 and 99??

see if u can get rid of the roots!

Does it look right? Going in the right direction maybe?

Oh, and I eliminated 7. and the sqrt of 63 is now the sqrt of 9

\(\sqrt {63} = \sqrt {9} \sqrt {7}=...?\)

Now it looks like \[2\sqrt{4x ^{3}}-\sqrt{9x ^{3}}+\sqrt{9}\]

\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...?\)
you just can't eliminate 7...

something is missing

But & goes into 7 one time and it goes into 63 9 times. Simplifying... Right?

but 7*

so, \(\sqrt {63}-\sqrt 7 = \sqrt{9}\sqrt{7}-\sqrt{7} = \sqrt{7}[3-1]=..?\)

you see what i did there ^ ?

No... :(

3x3

\(\sqrt {xy}=\sqrt{x}\sqrt{y}\)
ok ?
so, \(\sqrt{63}=\sqrt{9}\sqrt{7}=3 \sqrt{7}\)
right ?

ask if any doubts.

That does not look like any of my answer choices :(

I have no idea.

so, do you want me to start over ?
or take it from a particular step ?

Start from the beginning please?

right, lets see term by term,
so 63 is 9 times 7
so, \(\sqrt {63}= \sqrt {9}\sqrt{7}=...?\)

Yes. And what do we do with the \[\sqrt{7}\] just leave it where it is?

i was getting there...
you first tell me what will be \(\sqrt 9 \sqrt7 =... ?\)

simplify by dividing both \[\sqrt{7}\] and \[\sqrt{63}\] by 7?

dividing ? why ? no..

Oh... Grrr

lets finish 1 term entirely \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)

Yeah. 3.

so, \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)

\[\sqrt{3}\] \[\sqrt{2.6}??\]

since \(\sqrt 9=3\)
\(\sqrt {63}=\sqrt 9 \sqrt7 =3 \sqrt 7\)
got this ?

|dw:1359950980241:dw|

\[3\sqrt{7}\]

|dw:1359951141309:dw|

\(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=...?\)
ok with this ?

Wait...at the end multiplying \[\sqrt{7}\] by 2?

|dw:1359951264530:dw|

\(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=2 \sqrt 7\)

So now we have part of it completed?

yes, now take the terms with 'x'.

11 got into both

goes*

so, did you get this ?
\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}\)

44 = 4*11
x^3 = x*x^2

Why is there no 2?

in front of the \[\sqrt{44x^3}\]

Okay :)
\[4x \sqrt{11x}\]

similarly, what will be \(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)

Uh...?

Oh, I wasn't sure how to break it down. I see what you did now :)

\[x \sqrt{11x}+2\sqrt{7}\] :D

yes. thats correct .
i hope you got each step.

Yes! :D Haha thanks

welcome ^_^

this took a lot time :P

I'm slow, sorry xP