Question

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...

Answer 1

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

Answer 2

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

Answer 3

before combining see if you can simplify each term first

Answer 4

How so...?

Answer 5

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

Answer 6

see if u can get rid of the roots!

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

something is missing

Answer 14

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

Answer 15

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

Answer 16

but 7*

Answer 17

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

Answer 18

you see what i did there ^ ?

Answer 19

No... :(

Answer 20

@hartnn i see what he did

Answer 21

3x3

Answer 22

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

Answer 23

ask if any doubts.

Answer 24

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

Answer 25

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...

Answer 26

I have no idea.

Answer 27

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

Answer 28

Start from the beginning please?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

dividing ? why ? no..

Answer 34

Oh... Grrr

Answer 35

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

Answer 36

Yeah. 3.

Answer 37

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

Answer 38

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

Answer 39

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