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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

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

x = 0, x = - \frac{3 y}{1 + 2 y}; y \neq = - \frac{1}{2}

Solution steps

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

Write in the standard form

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

(1 + 2 y) x^{2} + 3 y x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 y \pm ( 3 y ) ^{2} - 4 ( 1 + 2 y ) \cdot 0}{2 ( 1 + 2 y )}

Simplify

(3 y)^{2} - 4 (1 + 2 y) \cdot 0 : 3 y

x_{1, 2} = \frac{- 3 y \pm 3 y}{2 ( 1 + 2 y )}; y \neq = - \frac{1}{2}

Separate the solutions

x_{1} = \frac{- 3 y + 3 y}{2 ( 1 + 2 y )}, x_{2} = \frac{- 3 y - 3 y}{2 ( 1 + 2 y )}

x = \frac{- 3 y + 3 y}{2 ( 1 + 2 y )} : 0

x = \frac{- 3 y - 3 y}{2 ( 1 + 2 y )} : - \frac{3 y}{1 + 2 y}

The solutions to the quadratic equation are:

x = 0, x = - \frac{3 y}{1 + 2 y}; y \neq = - \frac{1}{2}

2 t^{2} + 3 t - 1 = 0

(x^{2} + 3 x + 2) 2 - 8 (x^{2} + 3 x) = 4

2 x^{2} + 3 = 49

2 x^{2} + 3 = 2 x

7 x^{2} - 10 = x^{2} - 4

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