What are the solutions to the equation solvefor x,y=x^2-2x+1 ?

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

Find the zeros of solvefor x,y=x^2-2x+1

The zeros of solvefor x,y=x^2-2x+1 are x=1+sqrt(y),x=1-sqrt(y)

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solve for

x, y = x^{2} - 2 x + 1

x = 1 + y, x = 1 - y

Solution steps

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

Switch sides

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

Move

y

to the left side

x^{2} - 2 x + 1 - y = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot (1 - y) : 2 y

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + y, x = 1 - y

2 x^{2} + 4 = 2 (x + 2) - 3 x^{2}

\frac{( y + 1 ) ^{2}}{49} = 1

T^{1} = \frac{9}{5} T^{2} + 32

t^{2} - 4 t + 1 = 0

2 k (q + 1) (q - 1) + 2 q^{2} (n - k) = 0

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