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

The solutions to the equation solvefor x,xy^3=-3x^2y^4 are x=0,x=-1/(3y);\quad y\ne 0

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

The zeros of solvefor x,xy^3=-3x^2y^4 are x=0,x=-1/(3y);\quad y\ne 0

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solve for

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

x = 0, x = - \frac{1}{3 y}; y \neq = 0

Solution steps

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

Move

3 x^{2} y^{4}

to the left side

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

Factor

x y^{3} + 3 x^{2} y^{4} : x y^{3} (1 + 3 x y)

x y^{3} (1 + 3 x y) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 1 + 3 x y = 0

Solve

1 + 3 x y = 0 : x = - \frac{1}{3 y}; y \neq = 0

The solutions to the quadratic equation are:

x = 0, x = - \frac{1}{3 y}; y \neq = 0

3 r^{2} = 18

m^{2} = 225

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

x^{2} + x - 30 = 0

200 = 0.1 w^{2} + 2.155 w + 19.985

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