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# (root(9 + root(80)) + root(9 - root(80)))^2 =

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(root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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02 Jan 2009, 17:42
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$$(\sqrt{9+\sqrt{80}}+\sqrt{9-\sqrt{80}})^2=$$

A. 1
B. 9 - 4*5^1/2
C. 18 - 4*5^1/2
D. 18
E. 20
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 Jul 2015, 00:00, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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02 Jan 2009, 18:16
That's what I answered, but that is not the correct answer according to the GMATprep.
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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03 Jan 2009, 15:10
9 + sqrt(80) + 9 -sqrt(80) + 2 sqrt (81-80)

=20
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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03 Jan 2009, 20:55

thanks
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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04 Jan 2009, 17:54
These are the concept that you need to know to solve this problem

(x+y)^2 = x^2 +2xy+y^2
sqrtX times sqrt Y = sqrt XY
(a-b)(a+b) = a^2-b^2
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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05 Jan 2009, 16:55
(x+y)^2 = x^2+y^2+2xy
2*(9+sqrt(80))*(9-sqrt(80)=2*(81-80).

Rest of the explanation is clear from icandy.
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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10 Jan 2009, 10:46
squaring the term gives

9 + sqrt(80) + 9 - sqrt(80 + 2*(81 - 80) = 20
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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31 Aug 2009, 04:27
nagk78 wrote:
squaring the term gives

9 + sqrt(80) + 9 - sqrt(80 + 2*(81 - 80) = 20

I understand the concept of x^2 + y^2 +2xy ...

But can someone please explain how in the part 2xy they are getting 2*(81-80)?

Because 2xy is 2 * (sqrt(9) + sqrt(80) * sqrt(9) - sqrt(80)) ... or 2 * (sqrt(81)-sqrt(80))

i.e. not simply 81-80

**or is this one of those "find the closest approximation" questions.... and therefore should be spelled out in the question itself on the real GMAT?
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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31 Aug 2009, 14:54
$$(sqrt{a} + sqrt{b})^2$$ = a + b + $$2*sqrt{ab}$$

Here,
a = $$9 + sqrt{81}$$
b = $$9 - sqrt{81}$$

therefore,
answer = $$9 + [strike]sqrt{81}[/strike]$$ + $$9 - [strike]sqrt{81}[/strike]$$ + $$2*(81-80)$$
= 9 + 9 + 2 = 20
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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31 Aug 2009, 22:10
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i simply approximated

i know .... 4 sqrt(5) = 4 * 2.2 ~ 8.8

on simplyfying (a+b)^ 2 .....i get major terms as 2*9+2*...... = hence ans is more than 18 hence 20
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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14 Sep 2010, 23:30
im still not seeing the 81 and 80 here.....
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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14 Sep 2010, 23:36
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djslobo wrote:
im still not seeing the 81 and 80 here.....

Hi and welcome to Gmat Club. Below is properly formated solution for this problem:

You should know two properties:
1. $$(x+y)^2=x^2+2xy+y^2$$, ($$(x-y)^2=x^2-2xy+y^2$$);

2. $$(x+y)(x-y)=x^2-y^2$$.

$$(\sqrt{9+sqrt{80}}+\sqrt{9-\sqrt{80}})^2=(\sqrt{9+\sqrt{80}})^2+2(\sqrt{9+\sqrt{80}})(\sqrt{9-sqrt{80}})+(\sqrt{9-\sqrt{80}})^2=$$
$$=9+\sqrt{80}+2\sqrt{(9+\sqrt{80})(9-\sqrt{80})}+9-\sqrt{80}=18+2\sqrt{9^2-(\sqrt{80})^2}=18+2\sqrt{81-80}=18+2=20$$.

Hope it helps.
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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15 Sep 2010, 01:26
Bunuel wrote:
djslobo wrote:
im still not seeing the 81 and 80 here.....

Hi and welcome to Gmat Club. Below is properly formated solution for this problem:

You should know two properties:
1. $$(x+y)^2=x^2+2xy+y^2$$, ($$(x-y)^2=x^2-2xy+y^2$$);

2. $$(x+y)(x-y)=x^2-y^2$$.

$$(\sqrt{9+sqrt{80}}+\sqrt{9-\sqrt{80}})^2=(\sqrt{9+\sqrt{80}})^2+2(\sqrt{9+\sqrt{80}})(\sqrt{9-sqrt{80}})+(\sqrt{9-\sqrt{80}})^2=$$
$$=9+\sqrt{80}+2\sqrt{(9+\sqrt{80})(9-\sqrt{80})}+9-\sqrt{80}=18+2\sqrt{9^2-(\sqrt{80})^2}=18+2\sqrt{81-80}=18+2=20$$.

Hope it helps.

thnxs for the lucid explaination
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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28 Jul 2015, 13:14
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 = [#permalink]

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11 Sep 2016, 17:35
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