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

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

Find the zeros of 1=x^2+x

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1 = x^{2} + x

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

Decimal Notation

x = 0.61803 \dots, x = - 1.61803 \dots

Solution steps

1 = x^{2} + x

Switch sides

x^{2} + x = 1

Move

1

to the left side

x^{2} + x - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

1 = x^{2} + 1

4 x^{2} + 48 x + 144 = 0

x^{2} + 6 x = - 34

8 - 2 x^{2} = - 3

6 x^{2} + 7 x + 7 = 0

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