What are the solutions to the equation (x-3)^2-2(x-3)+1=0 ?

The solutions to the equation (x-3)^2-2(x-3)+1=0 are x=4

Find the zeros of (x-3)^2-2(x-3)+1=0

The zeros of (x-3)^2-2(x-3)+1=0 are x=4

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(x - 3)^{2} - 2 (x - 3) + 1 = 0

x = 4

Solution steps

(x - 3)^{2} - 2 (x - 3) + 1 = 0

Expand

(x - 3)^{2} - 2 (x - 3) + 1 : x^{2} - 8 x + 16

x^{2} - 8 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 16 = 0

x_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 1}

x = \frac{- ( - 8 )}{2 \cdot 1}

\frac{- ( - 8 )}{2 \cdot 1} = 4

x = 4

The solution to the quadratic equation is:

x = 4

- 0.001 x^{2} - 0.4 x + 210 = 0.0006 x^{2} + 0.26 x + 128

(x + 3)^{2} - 9 = 7

x^{2} + 6 x - 16 = 0

(1 - x)^{2} - 9 = 0

(x + 3)^{2} - 5 = 0

What are the solutions to the equation (x-3)^2-2(x-3)+1=0 ?
The solutions to the equation (x-3)^2-2(x-3)+1=0 are x=4
Find the zeros of (x-3)^2-2(x-3)+1=0
The zeros of (x-3)^2-2(x-3)+1=0 are x=4