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What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[

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What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 01 Nov 2010, 20:27
1
16
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A
B
C
D
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Difficulty:

  55% (hard)

Question Stats:

64% (01:45) correct 36% (02:07) wrong based on 539 sessions

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What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26

B. \(2\sqrt{29}\)

C. \(14 - 4\sqrt{5}\)

D. 14

E. \(14 + 4\sqrt{5}\)


Attachment:
solve.jpg
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 01 Nov 2010, 20:38
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What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26
B. \(2\sqrt{29}\)
C. \(14 - 4\sqrt{5}\)
D. 14
E. \(14 + 4\sqrt{5}\)


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=\)
\(=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 01 Nov 2010, 22:00
Bunuel wrote:
samark wrote:
Attachment:
solve.jpg


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=\)
\(=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.


That's right. Thanks Bunuel!
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 10 Mar 2015, 02:17
is there a quick way to solving these kind of questions? Even though I use the (x-y)^2 formula, it takes me a good 5 minutes..
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 10 Mar 2015, 05:10
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 02 Nov 2015, 11:46
1
Bunuel wrote:
What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26
B. \(2\sqrt{29}\)
C. \(14 - 4\sqrt{5}\)
D. 14
E. \(14 + 4\sqrt{5}\)


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.



Why are we using a^2 as square root of 49 and not as 7. How can I know that I need to keep one root as in 2[square_root]49-29

Instead of 2 (7 - [square_root]29) ???
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 20 Apr 2018, 08:03
Bunuel wrote:
What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26
B. \(2\sqrt{29}\)
C. \(14 - 4\sqrt{5}\)
D. 14
E. \(14 + 4\sqrt{5}\)


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=\)
\(=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.


I've been going over this and can't seem to wrap my head around this part:

(\sqrt{7+[square_root]29}[/square_root])^2 <---- isn't this another a^2 + b^2 + 2ab? why did this reduce to just 7+\sqrt{29}?
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 20 Apr 2018, 11:30
bp2013 wrote:
Bunuel wrote:
What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26
B. \(2\sqrt{29}\)
C. \(14 - 4\sqrt{5}\)
D. 14
E. \(14 + 4\sqrt{5}\)


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=\)
\(=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.


I've been going over this and can't seem to wrap my head around this part:

(\sqrt{7+[square_root]29}[/square_root])^2 <---- isn't this another a^2 + b^2 + 2ab? why did this reduce to just 7+\sqrt{29}?


\((\sqrt{x})^2=\sqrt{x}*\sqrt{x}=x\).

Similarly: \((\sqrt{7+\sqrt{29}})^2=\sqrt{7+\sqrt{29}}*\sqrt{7+\sqrt{29}}=7+\sqrt{29}\)
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Re: What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 20 Apr 2018, 12:13
Bunuel wrote:
bp2013 wrote:
Bunuel wrote:
What is the value of \((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2\)

A. -26
B. \(2\sqrt{29}\)
C. \(14 - 4\sqrt{5}\)
D. 14
E. \(14 + 4\sqrt{5}\)


You should apply two properties:
\((a-b)^2=a^2-2ab+b^2\);
\((a+b)(a-b)=a^2-b^2\)

\((\sqrt{7+\sqrt{29}}-\sqrt{7-\sqrt{29}})^2=(\sqrt{7+\sqrt{29}})^2-2(\sqrt{(7+\sqrt{29})(7-\sqrt{29})}+(\sqrt{7-\sqrt{29}})^2=\)
\(=7+\sqrt{29}-2\sqrt{49-29}+7-\sqrt{29}=14-2\sqrt{20}=14-4\sqrt{5}\).

Answer: C.


I've been going over this and can't seem to wrap my head around this part:

(\sqrt{7+[square_root]29}[/square_root])^2 <---- isn't this another a^2 + b^2 + 2ab? why did this reduce to just 7+\sqrt{29}?


\((\sqrt{x})^2=\sqrt{x}*\sqrt{x}=x\).

Similarly: \((\sqrt{7+\sqrt{29}})^2=\sqrt{7+\sqrt{29}}*\sqrt{7+\sqrt{29}}=7+\sqrt{29}\)


Thanks! I see my error. Since the sq rt covers the whole (7+29), that is to be treated as one term, NOT two, so it is not A^2 + 2ab + b^2, it is just that term multiplied by itself.
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What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[  [#permalink]

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New post 20 Apr 2018, 13:34
Bunuel way is the best approach but you can also use estimation if you are short on time. Looking at the answer choices, they are spread out far enough to estimate.

5^2 = 25 and 6^2 = 36. So estimate the sqrt29 to be 5.5.

Now 7 + 5.5 = 12.5 and 7 -5.5 = 1.5.
3^2 = 9 and 4^2 = 16. So estimate the sqrt12.5 to be 3.5 and Sqrt1.5 to be a little more than 1, or just assume 1.

Not you have (3.5 -1)^2 = 6.something

Looking at the answer choices, you can immediately cross of A (too small), D and E (too big).
Now you are left between b and C. If you notice b has sqrt 29, which you already assumed to be 5.5. Multiplying it by 2 will give you a number greater than 10. Answer C looks good because you have 14 - 4*2.something. This value will be closer to 6

Answer C
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What is the value of (\sqrt{7+\sqrt29}-\sqrt{7-[   [#permalink] 20 Apr 2018, 13:34
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