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Please help me solve? I'm looking at this problem and not seeing how to work it so that my answer comes out looking like the choices...

Mathematics
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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

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Other answers:

How so...?
Oh! like 11 goes into both 44 and 99??
see if u can get rid of the roots!
How would I go about doing that? Right now I have \[2\sqrt{4x ^{3}}-\sqrt{7}-\sqrt{9x ^{3}}+\sqrt{63}\]
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
\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...? \\\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)
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... :(
@hartnn i see what he did
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 :(
that was just simplification of constant terms.... \(3\sqrt 7- \sqrt 7= 2 \sqrt 7\) simplification of 'x' terms is still remaining. and more important thing is that you understand....so that you can do other problems on your own...
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|
ok, so now you have constant terms as \(3\sqrt 7-\sqrt 7\) can you combine this ? by factoring out \(\sqrt 7 \)
\(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=...?\) 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 \[\sqrt{44x^3}\]
there is 2 .. \(2\sqrt {44x^3} = 2\sqrt {4} \sqrt{x^2} \sqrt{11x}=2 \times 2x \sqrt {11x}= 4x \sqrt{11x}\)
Okay :) \[4x \sqrt{11x}\]
similarly, what will be \(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)
Uh...?
\(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=3x\sqrt {11x}\) because \(\sqrt 9=3, \: and \: \sqrt {x^2}=x\)
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

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