What are the solutions to the equation solvefor x,4x^2y^2+2xy-12=0 ?

The solutions to the equation solvefor x,4x^2y^2+2xy-12=0 are x= 3/(2y),x=-2/y ;\quad y\ne 0

Find the zeros of solvefor x,4x^2y^2+2xy-12=0

The zeros of solvefor x,4x^2y^2+2xy-12=0 are x= 3/(2y),x=-2/y ;\quad y\ne 0

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solve for

x, 4 x^{2} y^{2} + 2 x y - 12 = 0

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

Solution steps

4 x^{2} y^{2} + 2 x y - 12 = 0

Write in the standard form

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

4 y^{2} x^{2} + 2 y x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 y \pm ( 2 y ) ^{2} - 4 \cdot 4 y ^{2} ( - 12 )}{2 \cdot 4 y ^{2}}

Simplify

(2 y)^{2} - 4 \cdot 4 y^{2} (- 12) : 14 y

x_{1, 2} = \frac{- 2 y \pm 14 y}{2 \cdot 4 y ^{2}}; y \neq = 0

Separate the solutions

x_{1} = \frac{- 2 y + 14 y}{2 \cdot 4 y ^{2}}, x_{2} = \frac{- 2 y - 14 y}{2 \cdot 4 y ^{2}}

x = \frac{- 2 y + 14 y}{2 \cdot 4 y ^{2}} : \frac{3}{2 y}

x = \frac{- 2 y - 14 y}{2 \cdot 4 y ^{2}} : - \frac{2}{y}

The solutions to the quadratic equation are:

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

(x + 4)^{2} - (2 x + 3) (x - 2) = 42

6 (x + 1) - (4 - x) (x + 1) = 6 (4 - x)

y, 4 x^{3} - 2 x = 2 y^{2} + 7

λ^{2} - 4 = 0

3 x^{2} - 8 x + 5 = 0

What are the solutions to the equation solvefor x,4x^2y^2+2xy-12=0 ?
The solutions to the equation solvefor x,4x^2y^2+2xy-12=0 are x= 3/(2y),x=-2/y ;\quad y\ne 0
Find the zeros of solvefor x,4x^2y^2+2xy-12=0
The zeros of solvefor x,4x^2y^2+2xy-12=0 are x= 3/(2y),x=-2/y ;\quad y\ne 0