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(\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2

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(\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

A. 28
B. 29
C. 30
D. 31
E. 32

Reposting the question from another web, I do not understand the explanation. Please help with the details.
[Reveal] Spoiler: OA

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Last edited by obs23 on 06 Feb 2013, 23:49, edited 2 times in total.

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(\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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New post 06 Feb 2013, 23:38
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obs23 wrote:
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

A. 28
B. 29
C. 30
D. 31
E. 32

Reposting the question from another web, I do not understand the explanation. Please help with the details.

P.S. Please help with proper formatting - it does not seem to work as intended, but hope you get the idea.


Apply \((a+b)^2=a^2+2ab+b^2\):

\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2=\)
\(=(15+4\sqrt{14})+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}+(15-4\sqrt{14})=30+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}\).

Now, apply \((a+b)(a-b)=a^2-b^2\):

\(30+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=30+2\sqrt{15^2-16*14}=30+2*1=32\).

Answer: E.

Hope it's clear.
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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This answer is quite time consuming. Is there any other way to solve this problem?
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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Bunuel wrote:
obs23 wrote:
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

A. 28
B. 29
C. 30
D. 31
E. 32

Reposting the question from another web, I do not understand the explanation. Please help with the details.

P.S. Please help with proper formatting - it does not seem to work as intended, but hope you get the idea.


You can see that \(15+4\sqrt{14} = 15+2*2*\sqrt{2}\sqrt{7}\)

=\(2\sqrt{2}^2+\sqrt{7}^2+2*2*\sqrt{2}*\sqrt{7}\)

= \((2\sqrt{2}+\sqrt{7})^2\)

Similarly. the same for the other part and after the square root of the terms, you get \((2\sqrt{2}+\sqrt{7})\) + \((2\sqrt{2}-\sqrt{7})\) =\(2*2\sqrt{2}^2\) = 32

E.
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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Stiv wrote:
This answer is quite time consuming. Is there any other way to solve this problem?



I was able to solve it in 60 seconds

\(4\sqrt{14} = \sqrt{224}\)

\(15 = \sqrt{225}\)

\(= (\sqrt{\sqrt{225} + \sqrt{224}} + \sqrt{\sqrt{225} - \sqrt{224}})^2\)

\(= \sqrt{225} + \sqrt{224} + 2 \sqrt{(\sqrt{225} + \sqrt{224})( \sqrt{225} - \sqrt{224})} + \sqrt{225} - \sqrt{224}\)

= 15 + 2 + 15

= 32

Answer = E
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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The trick is knowing that anytime you have a number like \(200^2 - (199)(201)\) the answer will always be 1.

Proof:

\(x^2 - (x+1)(x-1) = x^2 - (x^2 - 1) = 1.\)

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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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put x=\(15+4\sqrt{14}\)
and y=\(15-4\sqrt{14}\)

==> \((\sqrt{x}+\sqrt{y})^2\)
==> \(x+2\sqrt{xy}+y\)
Substitute for x and y
===> \(30+2\sqrt{225-224}\)
===> 32
option D :)
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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New post 27 Oct 2016, 18:24
(a+b)^2 = a^2 + 2ab+ b^2

15+2+15 = 32

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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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obs23 wrote:
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

A. 28
B. 29
C. 30
D. 31
E. 32


We can let

x = √(15 + 4√14) and

y = √(15 - 4√14).

Since (x + y)^2 = x^2 + y^2 + 2xy, we have:

x^2 = (15 + 4√14)

y^2 = (15 - 4√14)

and

2xy is

2√(15 + 4√14) x √(15 - 4√14)

2√[(15 + 4√14)(15 - 4√14)]

Since (15 + 4√14)(15 - 4√14) can be written as a difference of squares, we have:

2√[(15^2) - (4√14)^2]

2√[225 - 224] = 2

Thus, the total value is:

(15 + 4√14) + (15 - 4√14) + 2 = 32

Answer: E
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2 [#permalink]

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New post 15 Nov 2017, 11:04
Once you reach this step:

\(2\sqrt{15^2-16*14}\).

You can further simplify \(16 * 14 = (15+1) (15-1) = 15^2 - 1^2\)

\(2\sqrt{15^2-(15^2 - 1^2)} = 2\)
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2   [#permalink] 15 Nov 2017, 11:04
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