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{[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2)

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{[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   Updated on: 23 Aug 2021, 06:57
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\(( \sqrt{8+\sqrt{63}}-\sqrt{8-\sqrt{63}})^2\)


(A) 14
(B) 14√2
(C) 16
(D) 16√2
(E) 18


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Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
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Originally posted by Carcass on 23 Aug 2021, 04:51.
Last edited by GeminiHeat on 23 Aug 2021, 06:57, edited 1 time in total.
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{[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   23 Aug 2021, 19:46
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\(( \sqrt{8+\sqrt{63}}-\sqrt{8-\sqrt{63}})^2\)

\(8 + \sqrt{63} + 8 - \sqrt{63} - 2(\sqrt{8+\sqrt{63}})(\sqrt{8-\sqrt{63}})\)

\(16 - 2\sqrt{(64 - 63)}\)

\(16 - 2\)

\(14\)

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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   23 Aug 2021, 13:56
can anyone give me detailed solution please?
thank you
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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   23 Aug 2021, 21:50
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The most obvious solution to follow, yet it is still elegant!!

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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   26 Oct 2021, 11:00
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arjunbir

Note: I refer the square root to sq.

Firstly, you can notice in the expression:
a= sq(8+sq63)
b= sq(8-sq63)
(a-b)^2 = a^2-2ab+b^2
By replacing the expression, it gives:
[sq(8+sq63) - sq(8-sq63)]^2 = 8+sq63 - 2[sq(8+sq63)][sq(8-sq63)] + 8-sq63
= 16 -2sq(8+sq63)sq(8-sq63)
sq(AB)=sqA*sqB. .
So, going back to our formula, it gives us:
= 16 - 2 sq[(8+sq63)(8-sq63)]
Then, we use the identity a^2-b^2=(a-b)(a+b)
= 16 - 2 sq(64 - 63)
= 16 - 2 sq1 = 16-2 = 14
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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink] New post   17 Apr 2024, 10:35
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink]
17 Apr 2024, 10:35
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