What are the solutions to the equation x=x(3-x) ?

The solutions to the equation x=x(3-x) are x=0,x=2

Find the zeros of x=x(3-x)

The zeros of x=x(3-x) are x=0,x=2

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x = x (3 - x)

x = 0, x = 2

Solution steps

x = x (3 - x)

Expand

x (3 - x) : 3 x - x^{2}

x = 3 x - x^{2}

Switch sides

3 x - x^{2} = x

Move

x

to the left side

- x^{2} + 2 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = 2

x = a (1 - \frac{z ^{2}}{b ^{2}}), z

x^{2} + 16 x = 41

(x - 2)^{2} + 2 = x

\frac{1}{3} π r^{2} (3) = 25 π

x, 4 (y - 2 x + 3) = (x + c)^{2}

What are the solutions to the equation x=x(3-x) ?
The solutions to the equation x=x(3-x) are x=0,x=2
Find the zeros of x=x(3-x)
The zeros of x=x(3-x) are x=0,x=2