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# If (2 - sqrt(5))x = -1, then x =

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If (2 - sqrt(5))x = -1, then x = [#permalink]

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25 Nov 2010, 15:17
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If $$(2-\sqrt{5})x = -1$$, then $$x=$$

A. 2 + sqrt(5)
B. 1 + (sqrt(5)/2
C. 1 - (sqrt(5)/2
D. 2 - sqrt(5)
E. -2 - sqrt(5)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 02 Jul 2013, 23:43, edited 1 time in total.
Renamed the topic and edited the question.
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Re: PS: if ( 2 - sqrt(5))x = -1, then x = [#permalink]

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25 Nov 2010, 15:28
vrajesh wrote:
if ( 2 - sqrt(5))x = -1, then x =

A. 2 + sqrt(5)
B. 1 + (sqrt(5)/2
C 1 - (sqrt(5)/2
D. 2 - sqrt(5)
E. -2 - sqrt(5)

Please explain. I can't seem to figure it out

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Question should be: If $$(2-\sqrt{5})x = -1$$, then $$x=$$

$$(2-\sqrt{5})x = -1$$ --> $$x=-\frac{1}{2-\sqrt{5}}=\frac{1}{\sqrt{5}-2}$$. Multiply both denominator and nominator by $$\sqrt{5}+2$$:

$$x=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}=\frac{\sqrt{5}+2}{\sqrt{5}^2-2^2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2$$.

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Re: PS: if ( 2 - sqrt(5))x = -1, then x = [#permalink]

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27 Nov 2010, 12:16
Geez, i feel stupid.... explain me plz the following stages:
1. x=-\frac{1}{2-\sqrt{5}}=\frac{1}{\sqrt{5}-2}
2. the last stage.

Thanks for ur time.
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Re: If (2 - sqrt(5))x = -1, then x = [#permalink]

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17 Jul 2014, 02:02
$$(2 - \sqrt{5}) x = -1$$

$$x = \frac{1}{\sqrt{5} - 2}$$

Multiple RHS numerator & denominator by $$\sqrt{5} + 2$$

$$x = \frac{\sqrt{5} + 2}{(\sqrt{5} + 2) (\sqrt{5} - 2)}$$

$$x = \frac{\sqrt{5} + 2}{5-4}$$

$$x = \sqrt{5} + 2$$

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Re: If (2 - sqrt(5))x = -1, then x =   [#permalink] 17 Jul 2014, 02:02
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# If (2 - sqrt(5))x = -1, then x =

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