0=x^2+x-1 — step-by-step solution

1) Let's solve in

ℝ

x^{2} + x - 1 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 1

and

c = -1

Δ = 1 - 4 \cdot 1 \cdot (-1)

Δ = 1 - 4 \cdot (-1)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-1)

4

Δ = 1 + 4

ⓘ Start from

1

and move forward

4

on the number line:

1 + 4

5

Δ = 5

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 1 - \sqrt{5}}{2 \cdot 1}

x_{1} = \frac{- 1 - \sqrt{5}}{2}

ⓘ The minus sign is factored out, changing the sign of each term:

- 1 - \sqrt{5}

- (1 + \sqrt{5})

x_{1} = - \frac{1 + \sqrt{5}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 1 + \sqrt{5}}{2 \cdot 1}

x_{2} = \frac{- 1 + \sqrt{5}}{2}

▶Solutions obtained:

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

2) The numeric value of:

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

is:

x = {\approx - \frac{1 + 2.2361}{2}, \approx \frac{2.2361 - 1}{2}}

x = {\approx - \frac{3.2361}{2}, \approx \frac{1.2361}{2}}

▷

x = {\approx -1.6180, \approx 0.6180}