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

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

Find the zeros of a^2-a=1

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

a^{2} - a = 1

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

Decimal Notation

a = 1.61803 \dots, a = - 0.61803 \dots

Solution steps

a^{2} - a = 1

Move

1

to the left side

a^{2} - a - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{1}{2} x^{2} - 4 x = 0

3 x^{2} - 3 x

5 + 22 z = 15 z^{2}

w (w + 4) = - 8

(a - 1)^{2} = 5 a

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