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

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Intern
Joined: 06 Feb 2013
Posts: 48

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Updated on: 06 Feb 2013, 23:49
5
33
00:00

Difficulty:

45% (medium)

Question Stats:

67% (01:59) correct 33% (02:14) wrong based on 566 sessions

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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.

Originally posted by obs23 on 06 Feb 2013, 23:13.
Last edited by obs23 on 06 Feb 2013, 23:49, edited 2 times in total.
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06 Feb 2013, 23:38
8
4
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 $$(a+b)(a-b)=a^2-b^2$$:

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

Hope it's clear.
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22 May 2014, 19:05
8
2
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

##### General Discussion
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19 Feb 2013, 02:41
1
This answer is quite time consuming. Is there any other way to solve this problem?
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Joined: 10 Oct 2012
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19 Feb 2013, 03:44
1
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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28 Aug 2015, 08:39
3
2
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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18 Dec 2015, 09:30
1
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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27 Oct 2016, 18:24
(a+b)^2 = a^2 + 2ab+ b^2

15+2+15 = 32
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12 Nov 2017, 07:29
1
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

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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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27 Jan 2020, 08:35
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Re: (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2   [#permalink] 27 Jan 2020, 08:35
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