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

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

Find the zeros of (1-x)(2+x)=1

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

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(1 - x) (2 + x) = 1

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

Decimal Notation

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

Solution steps

(1 - x) (2 + x) = 1

Expand

(1 - x) (2 + x) : 2 - x - x^{2}

2 - x - x^{2} = 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 ( - 1 ) \cdot 1}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{- 100}{- 100 - 3}

- 1 \leq m + 8

- 2 x < 12

3 s + 15 = 3 (s + 5)

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