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605-655 (Medium)|   Algebra|   Roots|      
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obs23
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 Updated on: 28 Jan 2024, 14:18
Originally posted by obs23 on 07 Feb 2013, 00:13.
Last edited by Bunuel on 28 Jan 2024, 14:18, edited 4 times in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

48% (02:02) correct 52% (02:07) wrong based on 1828 sessions

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

A. 28
B. 30
C. 32
D. 34
E. 36

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C
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 07 Feb 2013, 00:38
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obs23
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

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.
Show more

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: C.

Hope it's clear.
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 22 May 2014, 20:05
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Stiv
This answer is quite time consuming. Is there any other way to solve this problem?
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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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 19 Feb 2013, 03:41
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This answer is quite time consuming. Is there any other way to solve this problem?
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 19 Feb 2013, 04:44
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Bunuel
obs23
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

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.
Show more

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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 28 Aug 2015, 09:39
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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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 18 Dec 2015, 10:30
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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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 27 Oct 2016, 19:24
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(a+b)^2 = a^2 + 2ab+ b^2

15+2+15 = 32
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 12 Nov 2017, 08:29
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obs23
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

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

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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 15 Nov 2017, 12:04
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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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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 19 Mar 2023, 10:28
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↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧



We need to find the value of \((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

This is of the form \((a + b)^2\) and will be equal to \(a^2 + 2ab + b^2\)

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

Now, \((15+4\sqrt{14}) * (15-4\sqrt{14})\) is of the form (a+b) * (a-b) and will be equal to \(a^2 - b^2\)

=> 30 + 2 * \(\sqrt{ 15^2 - (4\sqrt{14})^2}\) = 30 + 2* \(\sqrt{225 - 16*14}\) = 30 + 2 * \(\sqrt{1}\) = 30 + 2 = 32

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Roots

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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 19 Mar 2023, 20:42
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Here is an approach which will help you to answer this in 30 seconds.
we have to use (a+b)^2 realise that a^2 + b^2 will give you 30(15+15). Now, we have +2ab ie an even number is to be added only one option is there of the type 30 + even number.
cheers!
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 13 Jan 2024, 06:42
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Bunuel, this one came in FE Mock 5. Please tag it.

Thank you
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 13 Jan 2024, 13:03
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AnkurGMAT20
Bunuel, this one came in FE Mock 5. Please tag it.

Thank you
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______________________
Added the tag. Thank you!
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 23 Jan 2024, 09:49
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Bunuel
obs23
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

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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can you explain how \(2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=2\sqrt{15^2-16*14}\) ?
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 23 Jan 2024, 09:56
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Bunuel
obs23
\((\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2\)

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.

can you explain how \(2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=2\sqrt{15^2-16*14}\) ?
Show more

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

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

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

    \(=2\sqrt{15^2 -(4\sqrt{14})^2}\)=
    \(=2\sqrt{15^2-16*14}\)
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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2 03 Apr 2025, 03:43
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Consider, \(15\) as \(a\) and \(4\sqrt{14}\) as \(b\) , upon simplifying you get, \(2a + 2(a^2-b^2) = 30 + 2(225-224) = 32\)
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