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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink]
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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]
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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]
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Re: {[(8+(63^2)]^(1/2) - [(8-(63^2)]^(1/2)}^(1/2) [#permalink]
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