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

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

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

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

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solve for

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

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

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(x + 3)^{2} + 6 (x + 3) + 5 = 0

- 2 x^{2} - 4 x - 1 = 0

x^{2} - 5 x + 13 = 11

p^{2} - 12 p = 0

2 t^{2} - 3 t - 1 = 0

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