What are the solutions to the equation 2x^2-4x+1=0 ?

The solutions to the equation 2x^2-4x+1=0 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

Find the zeros of 2x^2-4x+1=0

The zeros of 2x^2-4x+1=0 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

Decimal Notation

x = 1.70710 \dots, x = 0.29289 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 \cdot 1 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

\frac{r}{3} + \frac{2}{11} = \frac{39}{33}

17 - x < 18

2197

3 x^{2} + 13 x + 12

\frac{3}{7} x = - 9

What are the solutions to the equation 2x^2-4x+1=0 ?
The solutions to the equation 2x^2-4x+1=0 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2
Find the zeros of 2x^2-4x+1=0
The zeros of 2x^2-4x+1=0 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2