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555-605 (Medium)|   Algebra|   Roots|      
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pepo
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = Updated on: 12 Apr 2024, 02:08
Originally posted by pepo on 31 May 2016, 13:38.
Last edited by Bunuel on 12 Apr 2024, 02:08, edited 6 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)!

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35% (medium)

Question Stats:

74% (01:58) correct 26% (02:17) wrong based on 875 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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How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance ­

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Screenshot 2024-03-04 113739.png
Screenshot 2024-03-04 113739.png [ 191.44 KiB | Viewed 9859 times ]

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C
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KarishmaB
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 31 May 2016, 22:33
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pepo
\((\sqrt{15 + 4 \sqrt{14}} + \sqrt{15 - 4 \sqrt{14}})^2\)

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
Show more

The answer here needs to be (C) since the two terms are added in the expression.

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

Using \((A + B)^2 = A^2 + B^2 + 2AB\),

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

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

Now using \((A+B)(A - B) = A^2 - B^2\) under the square root sign, we get

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

= \(30 + 2 * 1\)

= \(32\)

Answer (C)
General Discussion
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Bunuel
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 31 May 2016, 14:40
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pepo
(\sqrt{15 + 4 [square_root]14}[/square_root] + \sqrt{15 - 4 [square_root]14}[/square_root])^2

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
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this is a copy of the following GMAT Prep question: root-9-root-80-root-9-root-95570.html
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 31 May 2016, 14:46
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pepo
(\sqrt{15 + 4 [square_root]14}[/square_root] + \sqrt{15 - 4 [square_root]14}[/square_root])^2

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
Show more
Dear pepo
I'm happy to respond. :-) If I understand correctly, this is the mathematical expression you were trying to convey:
Attachment:
root expression.JPG
root expression.JPG [ 10.43 KiB | Viewed 15143 times ]
Is this correct? It is not clear to me how to use the LaTex math notation to generate nested roots. Perhaps Bunuel or another expert can demonstrate that.

If I have the expression correct, this is a 100% valid question, a very clever question, very easy to do in one's head. It's important to know the fundamental polynomial patterns, the Square of a Difference and the Difference of Two Squares.

The square of the difference formula is
\((A - B)^2 = A^2 - 2AB + B^2\)
First look at the square of the individual terms.
\(A^2 = 15 + 4\sqrt{14}\)
\(B^2 = 15 - 4\sqrt{14}\)
When we add those, the radical expressions cancel, and we just get 30.

The cross term is more interesting. We combine the two radicals, and underneath the radicals, we get the difference of two squares pattern:
\((p + q)( p - q) = p^2 - q^2\)
Think about what's under the radical.
\((15 + 4\sqrt{14})(15 - 4\sqrt{14}) = 15^2 - (4^2)(14) = 15^2 - (16)(14)\)
We don't even have to do the calculation at the end: we simply need to recognize some advanced math factoring. From the difference of two squares,
\((K + 1)(K - 1) = K^2 - 1\)
For any integer K, the product of the integer one bigger (K + 1) and one smaller (K - 1) will always be one less than K^2. Without doing a calculation, we know that (16)(14) is exactly one less than 15^2. Thus, everything under the radical simplifies to 1. Thus
\(2AB = 2\)
and
\((A - B)^2 = A^2 - 2AB + B^2 = 30 - 2 = 28\)
Answer = (A)

Yes, this problem is full of material you should know well by test day. Please let me know if you have any questions.

Mike :-)
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 31 May 2016, 14:54
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mikemcgarry
pepo
(\sqrt{15 + 4 [square_root]14}[/square_root] + \sqrt{15 - 4 [square_root]14}[/square_root])^2

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
Dear pepo
I'm happy to respond. :-) If I understand correctly, this is the mathematical expression you were trying to convey:
Attachment:
root expression.JPG
Is this correct? It is not clear to me how to use the LaTex math notation to generate nested roots. Perhaps Bunuel or another expert can demonstrate that.

If I have the expression correct, this is a 100% valid question, a very clever question, very easy to do in one's head. It's important to know the fundamental polynomial patterns, the Square of a Difference and the Difference of Two Squares.

The square of the difference formula is
\((A - B)^2 = A^2 - 2AB + B^2\)
First look at the square of the individual terms.
\(A^2 = 15 + 4\sqrt{14}\)
\(B^2 = 15 - 4\sqrt{14}\)
When we add those, the radical expressions cancel, and we just get 30.

The cross term is more interesting. We combine the two radicals, and underneath the radicals, we get the difference of two squares pattern:
\((p + q)( p - q) = p^2 - q^2\)
Think about what's under the radical.
\((15 + 4\sqrt{14})(15 - 4\sqrt{14}) = 15^2 - (4^2)(14) = 15^2 - (16)(14)\)
We don't even have to do the calculation at the end: we simply need to recognize some advanced math factoring. From the difference of two squares,
\((K + 1)(K - 1) = K^2 - 1\)
For any integer K, the product of the integer one bigger (K + 1) and one smaller (K - 1) will always be one less than K^2. Without doing a calculation, we know that (16)(14) is exactly one less than 15^2. Thus, everything under the radical simplifies to 1. Thus
\(2AB = 2\)
and
\((A - B)^2 = A^2 - 2AB + B^2 = 30 - 2 = 28\)
Answer = (A)

Yes, this problem is full of material you should know well by test day. Please let me know if you have any questions.

Mike :-)
Show more

Edited the original post. Thank you.
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 01 Jun 2016, 03:34
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mikemcgarry
pepo
(\sqrt{15 + 4 [square_root]14}[/square_root] + \sqrt{15 - 4 [square_root]14}[/square_root])^2

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
Dear pepo
I'm happy to respond. :-) If I understand correctly, this is the mathematical expression you were trying to convey:
Attachment:
root expression.JPG
Is this correct? It is not clear to me how to use the LaTex math notation to generate nested roots. Perhaps Bunuel or another expert can demonstrate that.

If I have the expression correct, this is a 100% valid question, a very clever question, very easy to do in one's head. It's important to know the fundamental polynomial patterns, the Square of a Difference and the Difference of Two Squares.

The square of the difference formula is
\((A - B)^2 = A^2 - 2AB + B^2\)
First look at the square of the individual terms.
\(A^2 = 15 + 4\sqrt{14}\)
\(B^2 = 15 - 4\sqrt{14}\)
When we add those, the radical expressions cancel, and we just get 30.

The cross term is more interesting. We combine the two radicals, and underneath the radicals, we get the difference of two squares pattern:
\((p + q)( p - q) = p^2 - q^2\)
Think about what's under the radical.
\((15 + 4\sqrt{14})(15 - 4\sqrt{14}) = 15^2 - (4^2)(14) = 15^2 - (16)(14)\)
We don't even have to do the calculation at the end: we simply need to recognize some advanced math factoring. From the difference of two squares,
\((K + 1)(K - 1) = K^2 - 1\)
For any integer K, the product of the integer one bigger (K + 1) and one smaller (K - 1) will always be one less than K^2. Without doing a calculation, we know that (16)(14) is exactly one less than 15^2. Thus, everything under the radical simplifies to 1. Thus
\(2AB = 2\)
and
\((A - B)^2 = A^2 - 2AB + B^2 = 30 - 2 = 28\)
Answer = (A)

Yes, this problem is full of material you should know well by test day. Please let me know if you have any questions.

Mike :-)
Show more


In the given question, it is (A + B)^2 not (A - B)^2
:)
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pepo
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 01 Jun 2016, 08:11
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Thanks a lot for your help. Now it is much clear :)
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 16 Apr 2024, 15:10
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KarishmaB

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

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
The answer here needs to be (C) since the two terms are added in the expression.

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

Using \((A + B)^2 = A^2 + B^2 + 2AB\),

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

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

Now using \((A+B)(A - B) = A^2 - B^2\) under the square root sign, we get

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

= \(30 + 2 * 1\)

= \(32\)

Answer (C)
Show more
­
I dont understand this last step

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

= \(30 + 2 * 1\)

= \(32\)

how do we get 1 multiplied to 2 when there is so much else multiplied to it
 
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 17 Apr 2024, 01:12
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oneinabillion

KarishmaB

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

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

How much is probable to get a question similar to this one on test day?
What level do you think is this question?

Thanks in advance :)
The answer here needs to be (C) since the two terms are added in the expression.

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

Using \((A + B)^2 = A^2 + B^2 + 2AB\),

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

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

Now using \((A+B)(A - B) = A^2 - B^2\) under the square root sign, we get

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

= \(30 + 2 * 1\)

= \(32\)

Answer (C)
­
I dont understand this last step

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

= \(30 + 2 * 1\)

= \(32\)

how do we get 1 multiplied to 2 when there is so much else multiplied to it


 
Show more

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

\(­= 15 + 15 + 2*\sqrt{225 - 224}­= \)

\(­= 15 + 15 + 2*\sqrt{1}­= \)

\(­= 30 + 2 * 1­= \)

\(­= 32\)
Hope it's clear.­
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((15 - 4*14^(1/2))^(1/2) + (15 + 4*14^(1/2))^(1/2))^2 = 08 Jun 2025, 10:45
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