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

The solutions to the equation (x-3)^2=x-1 are x=5,x=2

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

The zeros of (x-3)^2=x-1 are x=5,x=2

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

x = 5, x = 2

Solution steps

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

Expand

(x - 3)^{2} : x^{2} - 6 x + 9

x^{2} - 6 x + 9 = x - 1

Move

1

to the left side

x^{2} - 6 x + 10 = x

Move

x

to the left side

x^{2} - 7 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 10 = 3

x_{1, 2} = \frac{- ( - 7 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 3}{2 \cdot 1}

x = \frac{- ( - 7 ) + 3}{2 \cdot 1} : 5

x = \frac{- ( - 7 ) - 3}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 5, x = 2

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

x^{2} - 28 x

x^{2} - 13 x

x^{2} - 15 x

v^{2} - 8 v + 45 = 0

What are the solutions to the equation (x-3)^2=x-1 ?
The solutions to the equation (x-3)^2=x-1 are x=5,x=2
Find the zeros of (x-3)^2=x-1
The zeros of (x-3)^2=x-1 are x=5,x=2