What are the solutions to the equation solvefor x,y^3+xy^2-4xy-4x^2=0 ?

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

Find the zeros of solvefor x,y^3+xy^2-4xy-4x^2=0

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

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solve for

x, y^{3} + x y^{2} - 4 x y - 4 x^{2} = 0

x = - y, x = \frac{y ^{2}}{4}

Solution steps

y^{3} + x y^{2} - 4 x y - 4 x^{2} = 0

Write in the standard form

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

- 4 x^{2} + (y^{2} - 4 y) x + y^{3} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( y ^{2} - 4 y ) \pm ( y ^{2} - 4 y ) ^{2} - 4 ( - 4 ) y ^{3}}{2 ( - 4 )}

Simplify

(y^{2} - 4 y)^{2} - 4 (- 4) y^{3} : y (y + 4)

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - y, x = \frac{y ^{2}}{4}

0 = x^{2} + \frac{5}{3} x - \frac{2}{3}

6 x^{2} - 5 x + 6 = 6 x + 3

(3 x - 2) (3 x + 2) = 77

6.53 t^{2} = 49

\frac{1}{10} x^{2} - 13 = - 3

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