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

The solutions to the equation x=(x-2)^2 are x=4,x=1

Find the zeros of x=(x-2)^2

The zeros of x=(x-2)^2 are x=4,x=1

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x = (x - 2)^{2}

x = 4, x = 1

Solution steps

x = (x - 2)^{2}

Expand

(x - 2)^{2} : x^{2} - 4 x + 4

x = x^{2} - 4 x + 4

Switch sides

x^{2} - 4 x + 4 = x

Move

x

to the left side

x^{2} - 5 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

x_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

x = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

x = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 4, x = 1

\frac{3 ^{5}}{3 ^{2}}

- 5 x^{2} + 9 + 7 x = 0

4 x^{2} - 16 x - 9

x^{2} - 3 x \geq 4

y^{2} - 50 = 0

What are the solutions to the equation x=(x-2)^2 ?
The solutions to the equation x=(x-2)^2 are x=4,x=1
Find the zeros of x=(x-2)^2
The zeros of x=(x-2)^2 are x=4,x=1