The square of (2 $x$ -1) is equal to 5. The values of $x$ are

x= (1 + √5)/2

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x= (-1 - √5)/2

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x= (-1 + √5)/2

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x= (1 + √6)/2

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Solution

The correct options are
A
x= (1 + √5)/2

D
x= (1 + √6)/2

$(2 x - 1)^{2}$ = 5

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

$2 x = \pm \sqrt{5} + 1$

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

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

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Similar questions

The square of (2 $x$ -1) is equal to 5. If one of the values of $x$ is $\frac{1 - \sqrt{5}}{2}$ , what is the value of $x$ ?

The square of (2 $x$ -1) is equal to 5. What is $x$ ?

(2 x + \frac{1}{x} + 1)

4 x^{2} + \frac{1}{x^{2}} + 5 + \frac{2}{x} + 4 x

\frac{3 x - 1}{4} - \frac{2 x + 5}{3} = \frac{5}{2} - 2 x,

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