What are the solutions to the equation solvefor y,x^3-xy+y^2=0 ?

The solutions to the equation solvefor y,x^3-xy+y^2=0 are y=(x+xsqrt(1-4x))/2 ,y=(x-xsqrt(1-4x))/2

Find the zeros of solvefor y,x^3-xy+y^2=0

The zeros of solvefor y,x^3-xy+y^2=0 are y=(x+xsqrt(1-4x))/2 ,y=(x-xsqrt(1-4x))/2

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solve for

y, x^{3} - x y + y^{2} = 0

y = \frac{x + x 1 - 4 x}{2}, y = \frac{x - x 1 - 4 x}{2}

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - x ) \pm ( - x ) ^{2} - 4 \cdot 1 \cdot x ^{3}}{2 \cdot 1}

Simplify

(- x)^{2} - 4 \cdot 1 \cdot x^{3} : x 1 - 4 x

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

Separate the solutions

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

y = \frac{- ( - x ) + x 1 - 4 x}{2 \cdot 1} : \frac{x + x 1 - 4 x}{2}

y = \frac{- ( - x ) - x 1 - 4 x}{2 \cdot 1} : \frac{x - x 1 - 4 x}{2}

The solutions to the quadratic equation are:

y = \frac{x + x 1 - 4 x}{2}, y = \frac{x - x 1 - 4 x}{2}

7 b^{2} + 14 b - 25 = - 4

x^{2} = 5^{2} + 6^{2}

3 x^{2} + 7 x + 6 = 0

4 x^{2} - 80 x + 356 = 0

(2 - 3 x)^{2} = x + 1

What are the solutions to the equation solvefor y,x^3-xy+y^2=0 ?
The solutions to the equation solvefor y,x^3-xy+y^2=0 are y=(x+xsqrt(1-4x))/2 ,y=(x-xsqrt(1-4x))/2
Find the zeros of solvefor y,x^3-xy+y^2=0
The zeros of solvefor y,x^3-xy+y^2=0 are y=(x+xsqrt(1-4x))/2 ,y=(x-xsqrt(1-4x))/2