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

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

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

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

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4 x^{2} - 2 x = 1

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

Decimal Notation

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

Solution steps

4 x^{2} - 2 x = 1

Move

1

to the left side

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}

4 x^{2} + 16 x + 16

\frac{3}{4} \cdot \frac{3}{4}

\frac{1}{4} t = 3

(- x + 2)^{3} (3 x + 1)^{4} (x - 2) (x + 3)^{8} > 0

6 w = - 12

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