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

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Joined: 08 Jun 2009
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Location: Newport Beach, CA
(root(9 + root(80)) + root(9 - root(80)))^2 =  [#permalink]

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Updated on: 22 Apr 2015, 03:52
2
22
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Difficulty:

35% (medium)

Question Stats:

67% (01:32) correct 33% (02:16) wrong based on 447 sessions

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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

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Originally posted by blakemancillas on 09 Jun 2010, 12:55.
Last edited by Bunuel on 22 Apr 2015, 03:52, edited 2 times in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 53069
Re: (root(9 + root(80)) + root(9 - root(80)))^2 =  [#permalink]

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09 Jun 2010, 13:38
9
5
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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Joined: 08 Nov 2010
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 =  [#permalink]

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28 Jan 2011, 22:35
thanks. good question.
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(root(9 + root(80)) + root(9 - root(80)))^2 =  [#permalink]

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20 Jul 2017, 07:32
Top Contributor
2
blakemancillas wrote:
$$(\sqrt{9+\sqrt{80}}+\sqrt{9-\sqrt{80}})^2=$$

A. 1
B. 9 - 4*5^1/2
C. 18 - 4*5^1/2
D. 18
E. 20

Notice that the expression is in the form (x + y)², where x = √(9 + √80) and y = √(9 - √80)

We know that (x + y)² = x² + 2xy + y²

If x = √(9 + √80), then x² = 9 + √80
If y = √(9 - √80), then y² = 9 - √80
Finally, xy = [√(9 + √80)][√(9 - √80)] = 81 - 80 = 1

So, we get:
(x + y)² = x² + 2xy + y²
= (9 + √80) + 2(1) + (9 - √80)
= 9 + √80 + 2 + 9 - √80
= 20

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

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05 Sep 2018, 08:19
blakemancillas wrote:
$$(\sqrt{9+\sqrt{80}}+\sqrt{9-\sqrt{80}})^2=$$

A. 1
B. 9 - 4*5^1/2
C. 18 - 4*5^1/2
D. 18
E. 20

$$A = \sqrt {9 + \sqrt {80} }$$
$$B = \sqrt {9 - \sqrt {80} }$$

$$?\,\,\,:\,\,\,{\text{expression}}\,\,{\text{ = }}\,\,{\left( {A + B} \right)^{\text{2}}}{\text{ = }}{{\text{A}}^{\text{2}}} + 2AB + {B^2}$$

$${A^2} = {\left( {\sqrt {9 + \sqrt {80} } } \right)^2} = 9 + \sqrt {80}$$
$${B^2} = {\left( {\sqrt {9 - \sqrt {80} } } \right)^2} = 9 - \sqrt {80}$$

$$AB = \sqrt {9 + \sqrt {80} } \cdot \sqrt {9 - \sqrt {80} } = \sqrt {\left( {9 + \sqrt {80} } \right)\left( {9 - \sqrt {80} } \right)} \,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\sqrt 1 = \boxed1$$
$$\left( * \right)\,\,\,\,\,\left( {9 + \sqrt {80} } \right)\left( {9 - \sqrt {80} } \right)\,\,\, = \,\,\,\,{9^2} - {\left( {\sqrt {80} } \right)^2} = 81 - 80 = 1$$

$$? = \left( {9 + \sqrt {80} } \right) + 2 \cdot \boxed1 + \left( {9 - \sqrt {80} } \right) = 20$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: (root(9 + root(80)) + root(9 - root(80)))^2 =   [#permalink] 05 Sep 2018, 08:19
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