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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ]

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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ] [#permalink] New post 23 Nov 2008, 21:53
I do not know how to type the code to get this to appear normal so I hope this is understandable:

SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ]

I have attached the problem in a word document as an image to make it easier to read!
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Re: Roots problem [#permalink] New post 23 Nov 2008, 22:02
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+4 or -4

the equation reduces to sqrt(16)

2*sqrt(63) = 2*sqrt(9*7) = 6sqrt(7)


2/(8+3*sqrt(7) ) = (2 * (8-3*sqrt(7)))/(8^2 - (3sqrt(7))^2)
= 16 - 6*sqrt(7)
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Re: Roots problem [#permalink] New post 24 Nov 2008, 01:28
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Answer is 16

After substituting 2sqrt63 with 6sqrt7, we should multify 2/(8+3sqrt7) by (8-3sqrt7). We get 6sqrt7 + (16-6sqrt7)/(64-63) = 16
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Re: Roots problem [#permalink] New post 24 Nov 2008, 08:10
I understand the 6(sqrt7) but not sure why we should multiply 2/(8+3sqrt7) by (8-3sqrt7)... why the MINUS...

THANKS ALL FOR YOUR HELP!!
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Re: Roots problem [#permalink] New post 24 Nov 2008, 10:59
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hardaway7, a^2-b^2 = (a-b)(a+b) We should multify 8+3sqrt7 by 8-3sqrt7 to get 64-63=1

emailsector, igon, may be you don't understand the question as it is written, then please download word file attached to it. Answer is 16
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Re: Roots problem [#permalink] New post 24 Nov 2008, 13:12
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hardaway7 wrote:
I understand the 6(sqrt7) but not sure why we should multiply 2/(8+3sqrt7) by (8-3sqrt7)... why the MINUS...

THANKS ALL FOR YOUR HELP!!


Whenever you see a root in the denominator, the first instinct should be to get rid of it. Its easy to get rid of (a + Vb) or (a -Vb) in the denominator, since you can simply use the a^2-b^2 formula, and hence bring your root to the numerator, where its easier to manage.

1/(a+Vb)
= (a-Vb)/{(a+Vb)(a-Vb)} .... multiplying both numerator and denominator by (a-Vb)
= (a -Vb) / a^2 - b
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Re: Roots problem [#permalink] New post 24 Nov 2008, 23:03
Wow thanks- I read more about this after your comments - After all the practice I have done I don't know why I haven't come across this often enough to remember! THANKS!

If anyone else read this problem by chance and didn't understand - I have attached a word document on rationalize denominators that I found helpful
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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ] [#permalink] New post 03 Feb 2015, 14:05
Are we sure about these responses?
I also have the question:

√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7

And according to the answers the correct one should be D, so 4... And it is 4, because I used the calculator to check it.
Unfortunatelly, I cannot seem to be able to solve it on my own...

Perhaps, those of you that said that the answer is 16, you forgot to take the square root?
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Re: SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ] [#permalink] New post 03 Feb 2015, 14:08
Expert's post
pacifist85 wrote:
Are we sure about these responses?
I also have the question:

√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7

And according to the answers the correct one should be D, so 4... And it is 4, because I used the calculator to check it.
Unfortunatelly, I cannot seem to be able to solve it on my own...

Perhaps, those of you that said that the answer is 16, you forgot to take the square root?


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Re: SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ]   [#permalink] 03 Feb 2015, 14:08
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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ]

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