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

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

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

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

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solve for

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

y = - x + 1, y = - x - 1

Solution steps

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

Move

1

to the left side

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

9 x^{2} = 54 x - 83

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

x^{2} + \frac{1}{2} x - 4

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

- x^{2} + 10 x = 16

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