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(roots of 15+4 roots of 14 +.....

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(roots of 15+4 roots of 14 +.....  [#permalink]

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New post Updated on: 31 May 2016, 13:53
5
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:37) correct 28% (02:11) wrong based on 139 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

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 :)

Originally posted by pepo on 31 May 2016, 12:38.
Last edited by Bunuel on 31 May 2016, 13:53, edited 1 time in total.
Edited the question.
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 31 May 2016, 13:40
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 31 May 2016, 13:46
pepo wrote:
(\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
root expression.JPG [ 10.43 KiB | Viewed 1935 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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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 31 May 2016, 13:54
mikemcgarry wrote:
pepo wrote:
(\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 :-)


Edited the original post. Thank you.
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 31 May 2016, 21:33
pepo wrote:
\((\sqrt{15 + 4 \sqrt{14}} + \sqrt{15 - 4 \sqrt{14}})^2\)

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)
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 01 Jun 2016, 02:34
mikemcgarry wrote:
pepo wrote:
(\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 :-)



In the given question, it is (A + B)^2 not (A - B)^2
:)
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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New post 01 Jun 2016, 07:11
Thanks a lot for your help. Now it is much clear :)
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Re: (roots of 15+4 roots of 14 +.....  [#permalink]

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Re: (roots of 15+4 roots of 14 +.....   [#permalink] 08 Nov 2017, 12:20
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