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

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Current Student
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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ] [#permalink]  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!
Attachments

problem.doc [23 KiB]

Intern
Joined: 04 Jun 2008
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Re: Roots problem [#permalink]  23 Nov 2008, 22:02
2
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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]  24 Nov 2008, 01:28
1
KUDOS

After substituting 2sqrt63 with 6sqrt7, we should multify 2/(8+3sqrt7) by (8-3sqrt7). We get 6sqrt7 + (16-6sqrt7)/(64-63) = 16
Current Student
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Re: Roots problem [#permalink]  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...

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Re: Roots problem [#permalink]  24 Nov 2008, 10:59
1
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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]  24 Nov 2008, 13:12
1
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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...

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
Current Student
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Re: Roots problem [#permalink]  24 Nov 2008, 23:03

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
Attachments

Rationalizing Denominators.doc [49.5 KiB]

Senior Manager
Status: Math is psycho-logical
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SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ] [#permalink]  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]  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?

Discussed here: topic-101735.html
_________________
Re: SQRT[ 2(SQRT of 63) + 2/8+3(SQRT of 7) ]   [#permalink] 03 Feb 2015, 14:08
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