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

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vrajesh
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If (2 - sqrt(5))x = -1, then x = [#permalink] New post   Updated on: 03 Jul 2013, 00:43
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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)
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A

Originally posted by vrajesh on 25 Nov 2010, 16:17.
Last edited by Bunuel on 03 Jul 2013, 00:43, 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] New post   25 Nov 2010, 16:28
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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


Please read and follow:
writing-mathematical-symbols-in-posts-72468.html
how-to-improve-the-forum-search-function-for-others-99451.html

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 \(\sqrt{5}+2\):

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

Answer: A.
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Re: If (2 - sqrt(5))x = -1, then x = [#permalink] New post   18 Feb 2021, 07:30
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(2 - sqrt(5))x = -1

Let's multiply both sides by (2 + sqrt(5)):

(2 - sqrt(5)) (2 + sqrt(5))x = -1(2 + sqrt(5))

(4-5)x = -1(2 + sqrt(5))

x = (2 + sqrt(5))
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Re: PS: if ( 2 - sqrt(5))x = -1, then x = [#permalink] New post   27 Nov 2010, 13:16
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] New post   17 Jul 2014, 03:02
\((2 - \sqrt{5}) x = -1\)

\(x = \frac{1}{\sqrt{5} - 2}\)

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

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

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

\(x = \sqrt{5} + 2\)

Answer = A
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Re: If (2 - sqrt(5))x = -1, then x = [#permalink] New post   18 Feb 2021, 07:59
vrajesh wrote:
If \((2-\sqrt{5})x = -1\), then \(x=\)

A. \(2 + \sqrt{5}\)
B. \(\frac{1 + \sqrt{5}}{2}\)
C. \(\frac{1 - \sqrt{5}}{2}\)
D. \(2 - \sqrt{5}\)
E. \(-2 - \sqrt{5}\)


\((2-\sqrt{5})x = -1\)

\(x = \frac{-1}{(2-\sqrt{5})}\)

\(x = \frac{-1}{(2-\sqrt{5})} * \frac{(2+\sqrt{5})}{(2+\sqrt{5})}\)

\(x = \frac{-1 (2+\sqrt{5})}{(4-5)}\)

\(x = \frac{-1 (2+\sqrt{5})}{(-1)}\)

\(x = 2+\sqrt{5}\)

Ans A
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Re: If (2 - sqrt(5))x = -1, then x = [#permalink] New post   17 Aug 2021, 04:48
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)


\((2-\sqrt{5})x = -1\)
\(x =\frac{ -1}{(2-\sqrt{5})}\)

Rationalizing the denominator, we get
\(x =\frac{ -1(2+\sqrt{5})}{(4-5)}\) \(= 2+\sqrt{5}\)

Hence, OA is (A).
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If (2 - sqrt(5))x = -1, then x = [#permalink] New post   04 Apr 2023, 10:26
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↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧




Given that \((2-\sqrt{5})x\) = -1 and we need to find the value of x

\((2-\sqrt{5})x\) = -1
=> x = \(\frac{-1}{2-\sqrt{5} }\) = \(\frac{1 }{ \sqrt{5} - 2}\)

Multiplying numerator and denominator with \(\sqrt{5} + 2\) we get

x = \(\frac{1 * \sqrt{5} + 2}{ (\sqrt{5} - 2)*(\sqrt{5} + 2)}\)
= \(\frac{\sqrt{5} + 2}{(\sqrt{5})^2 - 2^2}\) = \(\frac{\sqrt{5} + 2}{5 - 4}\) = \(\sqrt{5} + 2\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Rationalize Roots

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If (2 - sqrt(5))x = -1, then x = [#permalink]
04 Apr 2023, 10:26
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