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

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

Find the zeros of x-1=(2-x)^2

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

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x - 1 = (2 - x)^{2}

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

Solution steps

x - 1 = (2 - x)^{2}

Expand

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

x - 1 = 4 - 4 x + x^{2}

Switch sides

4 - 4 x + x^{2} = x - 1

Move

1

to the left side

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

Move

x

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

5 b^{2} k^{2} + 25 b k^{2} - 250 k^{2}

6 = x - 7

((3^{2} 2^{4})^{5} (1 2^{5} 1 8^{30})^{4})^{- 2}

102 = x 2

a + b - \frac{a - 3}{3}

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