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

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

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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(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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vrajesh
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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vrajesh
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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↧↧↧ 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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