P.nut1996
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...
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P.nut1996
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\[2\sqrt{44x ^{3}}-\sqrt{7}-\sqrt{99x ^{3}}+\sqrt{63}\]
P.nut1996
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Combine like terms, right?
I don't see how to combine \[2\sqrt{44x ^{3}}\] and \[-\sqrt{99x ^{3}}\]
javawarrior
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before combining see if you can simplify each term first
P.nut1996
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How so...?
P.nut1996
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Oh! like 11 goes into both 44 and 99??
javawarrior
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see if u can get rid of the roots!
P.nut1996
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How would I go about doing that? Right now I have \[2\sqrt{4x ^{3}}-\sqrt{7}-\sqrt{9x ^{3}}+\sqrt{63}\]
P.nut1996
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Does it look right? Going in the right direction maybe?
P.nut1996
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Oh, and I eliminated 7. and the sqrt of 63 is now the sqrt of 9
hartnn
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\(\sqrt {63} = \sqrt {9} \sqrt {7}=...?\)
P.nut1996
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Now it looks like \[2\sqrt{4x ^{3}}-\sqrt{9x ^{3}}+\sqrt{9}\]
hartnn
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\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...?\)
you just can't eliminate 7...
Andresfon12
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something is missing
hartnn
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\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...? \\\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)
P.nut1996
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But & goes into 7 one time and it goes into 63 9 times. Simplifying... Right?
P.nut1996
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but 7*
hartnn
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so, \(\sqrt {63}-\sqrt 7 = \sqrt{9}\sqrt{7}-\sqrt{7} = \sqrt{7}[3-1]=..?\)
hartnn
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you see what i did there ^ ?
P.nut1996
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No... :(
Andresfon12
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@hartnn i see what he did
Andresfon12
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3x3
hartnn
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\(\sqrt {xy}=\sqrt{x}\sqrt{y}\)
ok ?
so, \(\sqrt{63}=\sqrt{9}\sqrt{7}=3 \sqrt{7}\)
right ?
hartnn
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ask if any doubts.
P.nut1996
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That does not look like any of my answer choices :(
hartnn
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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...
P.nut1996
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I have no idea.
hartnn
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so, do you want me to start over ?
or take it from a particular step ?
P.nut1996
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Start from the beginning please?
hartnn
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right, lets see term by term,
so 63 is 9 times 7
so, \(\sqrt {63}= \sqrt {9}\sqrt{7}=...?\)
P.nut1996
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Yes. And what do we do with the \[\sqrt{7}\] just leave it where it is?
hartnn
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i was getting there...
you first tell me what will be \(\sqrt 9 \sqrt7 =... ?\)
P.nut1996
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simplify by dividing both \[\sqrt{7}\] and \[\sqrt{63}\] by 7?
hartnn
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dividing ? why ? no..
P.nut1996
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Oh... Grrr
hartnn
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lets finish 1 term entirely \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)
P.nut1996
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Yeah. 3.
hartnn
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so, \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)
P.nut1996
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\[\sqrt{3}\] \[\sqrt{2.6}??\]
hartnn
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since \(\sqrt 9=3\)
\(\sqrt {63}=\sqrt 9 \sqrt7 =3 \sqrt 7\)
got this ?
mathstudent55
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|dw:1359950980241:dw|
P.nut1996
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\[3\sqrt{7}\]
mathstudent55
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|dw:1359951141309:dw|
hartnn
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ok, so now you have constant terms as \(3\sqrt 7-\sqrt 7\)
can you combine this ? by factoring out \(\sqrt 7 \)
hartnn
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\(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=...?\)
ok with this ?
P.nut1996
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Wait...at the end multiplying \[\sqrt{7}\] by 2?
mathstudent55
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|dw:1359951264530:dw|
hartnn
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\(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=2 \sqrt 7\)
P.nut1996
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So now we have part of it completed?
hartnn
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yes, now take the terms with 'x'.
P.nut1996
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11 got into both
P.nut1996
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goes*
hartnn
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so, did you get this ?
\(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}\)
hartnn
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44 = 4*11
x^3 = x*x^2
P.nut1996
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Why is there no 2?
P.nut1996
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in front of the \[\sqrt{44x^3}\]
hartnn
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there is 2 ..
\(2\sqrt {44x^3} = 2\sqrt {4} \sqrt{x^2} \sqrt{11x}=2 \times 2x \sqrt {11x}= 4x \sqrt{11x}\)
P.nut1996
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Okay :)
\[4x \sqrt{11x}\]
hartnn
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similarly, what will be \(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)
P.nut1996
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Uh...?
hartnn
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\(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=3x\sqrt {11x}\)
because \(\sqrt 9=3, \: and \: \sqrt {x^2}=x\)
P.nut1996
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Oh, I wasn't sure how to break it down. I see what you did now :)
P.nut1996
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\[x \sqrt{11x}+2\sqrt{7}\] :D
hartnn
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yes. thats correct .
i hope you got each step.
P.nut1996
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Yes! :D Haha thanks
hartnn
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welcome ^_^
Andresfon12
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this took a lot time :P
P.nut1996
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I'm slow, sorry xP