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

The solutions to the equation a=2a(1-a) are a=0,a= 1/2

Find the zeros of a=2a(1-a)

The zeros of a=2a(1-a) are a=0,a= 1/2

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a = 2 a (1 - a)

a = 0, a = \frac{1}{2}

Solution steps

a = 2 a (1 - a)

Expand

2 a (1 - a) : 2 a - 2 a^{2}

a = 2 a - 2 a^{2}

Switch sides

2 a - 2 a^{2} = a

Move

a

to the left side

- 2 a^{2} + a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 0 = 1

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

Separate the solutions

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

a = \frac{- 1 + 1}{2 ( - 2 )} : 0

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

The solutions to the quadratic equation are:

a = 0, a = \frac{1}{2}

e^{\frac{πi}{2}}

x^{2} + 2 x + 1 - p^{2}

z^{2} + \frac{5}{2} z + 1 = 0

i^{32}

7 n = 2 n - 5 (- 5)

What are the solutions to the equation a=2a(1-a) ?
The solutions to the equation a=2a(1-a) are a=0,a= 1/2
Find the zeros of a=2a(1-a)
The zeros of a=2a(1-a) are a=0,a= 1/2