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

The solutions to the equation quadraticformula x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)

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

The zeros of quadraticformula x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

x = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

w^{2} + 6 w - 11 = 0

x^{2} - 4 x + 3 = 0

- 9 x^{2} = 0

x^{2} + x = \frac{11}{4}

5 x^{2} + 3 x + 4 = 0

What are the solutions to the equation quadraticformula x^2-4x+1=0 ?
The solutions to the equation quadraticformula x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)
Find the zeros of quadraticformula x^2-4x+1=0
The zeros of quadraticformula x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)