What are the solutions to the equation x^2-2x+1=4 ?

The solutions to the equation x^2-2x+1=4 are x=3,x=-1

Find the zeros of x^2-2x+1=4

The zeros of x^2-2x+1=4 are x=3,x=-1

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

x = 3, x = - 1

Solution steps

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

Move

4

to the left side

x^{2} - 2 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

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

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 4}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 3, x = - 1

\frac{3}{11}

- x^{4} + 2 x^{3} - 2 x + 1

2 (2) + (8) = - 4

\frac{6 c ^{2} + 13 c - 63}{6 c ^{2} - 17 c + 7} \cdot \frac{2 c ^{2} - 9 c + 4}{12 c + 54}

- 19 \leq 3 x - 5 \leq 1

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