What are the solutions to the equation solvefor y,x^2+y^2-2xy-1=0 ?

The solutions to the equation solvefor y,x^2+y^2-2xy-1=0 are y=x+1,y=x-1

Find the zeros of solvefor y,x^2+y^2-2xy-1=0

The zeros of solvefor y,x^2+y^2-2xy-1=0 are y=x+1,y=x-1

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solve for

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

y = x + 1, y = x - 1

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 x ) \pm ( - 2 x ) ^{2} - 4 \cdot 1 \cdot ( x ^{2} - 1 )}{2 \cdot 1}

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

y_{1, 2} = \frac{- ( - 2 x ) \pm 2}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 x ) + 2}{2 \cdot 1}, y_{2} = \frac{- ( - 2 x ) - 2}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

y = x + 1, y = x - 1

0 = - 12 x^{2} - 24 x

5 x^{2} - 28 x = 12

5 x^{2} - 4 x = 2

27 x^{2} + 9 x + 3 = 0

2 (y - 2)^{2} = 150

What are the solutions to the equation solvefor y,x^2+y^2-2xy-1=0 ?
The solutions to the equation solvefor y,x^2+y^2-2xy-1=0 are y=x+1,y=x-1
Find the zeros of solvefor y,x^2+y^2-2xy-1=0
The zeros of solvefor y,x^2+y^2-2xy-1=0 are y=x+1,y=x-1