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

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Re: If (2 - sqrt(5))x = -1, then x = [#permalink]
$$(2 - \sqrt{5}) x = -1$$

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

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]
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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: If (2 - sqrt(5))x = -1, then x = [#permalink]
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]
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]
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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 = $$\frac{1 * \sqrt{5} + 2}{ (\sqrt{5} - 2)*(\sqrt{5} + 2)}$$
= $$\frac{\sqrt{5} + 2}{(\sqrt{5})^2 - 2^2}$$ = $$\frac{\sqrt{5} + 2}{5 - 4}$$ = $$\sqrt{5} + 2$$