What are the solutions to the equation 2x=4x^2-1 ?

The solutions to the equation 2x=4x^2-1 are x=(1+sqrt(5))/4 ,x=(1-sqrt(5))/4

Find the zeros of 2x=4x^2-1

The zeros of 2x=4x^2-1 are x=(1+sqrt(5))/4 ,x=(1-sqrt(5))/4

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2 x = 4 x^{2} - 1

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

Decimal Notation

x = 0.80901 \dots, x = - 0.30901 \dots

Solution steps

2 x = 4 x^{2} - 1

Switch sides

4 x^{2} - 1 = 2 x

Move

2 x

to the left side

4 x^{2} - 1 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 2 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 4}

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

x = \frac{- ( - 2 ) - 2 5}{2 \cdot 4} : \frac{1 - 5}{4}

The solutions to the quadratic equation are:

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

\frac{4}{9} a^{2} - \frac{1}{4} a = 5

(5 + y)^{2} + y^{2} = 37

x^{2} + 6 x = - 5

x, f = 2 x^{2} - 2 y

- 16 t^{2} + 1600 = 0

What are the solutions to the equation 2x=4x^2-1 ?
The solutions to the equation 2x=4x^2-1 are x=(1+sqrt(5))/4 ,x=(1-sqrt(5))/4
Find the zeros of 2x=4x^2-1
The zeros of 2x=4x^2-1 are x=(1+sqrt(5))/4 ,x=(1-sqrt(5))/4