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

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

Find the zeros of x=1-x^2

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

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x = 1 - x^{2}

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

Solution steps

x = 1 - x^{2}

Switch sides

1 - x^{2} = x

Move

x

to the left side

1 - x^{2} - x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 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}

2 w^{3} - 8 w^{2} - 90 w

- 6 r + 3 = - 39

(\frac{m ^{- 1} m ^{5}}{m ^{- 2}})^{- 3}

x^{2} + (\frac{1}{6})^{2} = 1^{2}

4 x^{3} + 35 x^{2} + 88 x + 48

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