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...

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

before combining see if you can simplify each term first

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## More answers

- anonymous

How so...?

- anonymous

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

- anonymous

see if u can get rid of the roots!

- anonymous

How would I go about doing that? Right now I have \[2\sqrt{4x ^{3}}-\sqrt{7}-\sqrt{9x ^{3}}+\sqrt{63}\]

- anonymous

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

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

something is missing

- hartnn

\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...? \\\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)

- anonymous

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

- anonymous

but 7*

- hartnn

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

- hartnn

you see what i did there ^ ?

- anonymous

No... :(

- anonymous

@hartnn i see what he did

- anonymous

3x3

- hartnn

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

- hartnn

ask if any doubts.

- anonymous

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

- hartnn

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...

- anonymous

I have no idea.

- hartnn

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

- anonymous

Start from the beginning please?

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

dividing ? why ? no..

- anonymous

Oh... Grrr

- hartnn

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

- anonymous

Yeah. 3.

- hartnn

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

- anonymous

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

- hartnn

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

- mathstudent55

|dw:1359950980241:dw|

- anonymous

\[3\sqrt{7}\]

- mathstudent55

|dw:1359951141309:dw|

- hartnn

ok, so now you have constant terms as \(3\sqrt 7-\sqrt 7\)
can you combine this ? by factoring out \(\sqrt 7 \)

- hartnn

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

- anonymous

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

- mathstudent55

|dw:1359951264530:dw|

- hartnn

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

- anonymous

So now we have part of it completed?

- hartnn

yes, now take the terms with 'x'.

- anonymous

11 got into both

- anonymous

goes*

- hartnn

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

- hartnn

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

- anonymous

Why is there no 2?

- anonymous

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

- hartnn

there is 2 ..
\(2\sqrt {44x^3} = 2\sqrt {4} \sqrt{x^2} \sqrt{11x}=2 \times 2x \sqrt {11x}= 4x \sqrt{11x}\)

- anonymous

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

- hartnn

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

- anonymous

Uh...?

- hartnn

\(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=3x\sqrt {11x}\)
because \(\sqrt 9=3, \: and \: \sqrt {x^2}=x\)

- anonymous

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

- anonymous

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

- hartnn

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

- anonymous

Yes! :D Haha thanks

- hartnn

welcome ^_^

- anonymous

this took a lot time :P

- anonymous

I'm slow, sorry xP

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