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

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

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

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

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solve for

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

y = - x

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(2 x)^{2} - 4 \cdot 1 \cdot x^{2} = 0

y_{1, 2} = \frac{- 2 x \pm 0}{2 \cdot 1}

y = \frac{- 2 x}{2 \cdot 1}

Simplify

\frac{- 2 x}{2 \cdot 1} : - x

y = - x

The solution to the quadratic equation is:

y = - x

- 4 x^{2} - 24 x - 36 = 0

x^{2} - 109 = - 8 x

3 (x + 7)^{2} - 32 = - 29

x^{2} + 5 x - 14 = 0

6 b^{2} - 7 b - 21 = - 1

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