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

The solutions to the equation x=x^2-2x+1 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

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

The zeros of x=x^2-2x+1 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

x = 2.61803 \dots, x = 0.38196 \dots

Solution steps

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

Switch sides

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

Move

x

to the left side

x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : \frac{3 + 5}{2}

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : \frac{3 - 5}{2}

The solutions to the quadratic equation are:

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

9 x^{2} + 10 x = 5

x, - 2 x^{2} + y^{3} - 7 y + x = 0

5^{2} = a^{2} + (\frac{5}{2})^{2}

r, (2 - r)^{2} + (4 - r)^{2} = (r + 2)^{2}

5.7614 a^{2} + 2593 a - 516461.55 = 0

What are the solutions to the equation x=x^2-2x+1 ?
The solutions to the equation x=x^2-2x+1 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2
Find the zeros of x=x^2-2x+1
The zeros of x=x^2-2x+1 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2