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

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

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

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

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(x - 2)^{2} = x

x = 4, x = 1

Solution steps

(x - 2)^{2} = x

Expand

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

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

20 < \frac{z}{4}

f = - (\frac{1}{2})^{2} + 3 (\frac{1}{2}) + 1

x^{2} - 289

- 3 x - 9 = 0

5 - 6 x \leq - 20

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