What are the solutions to the equation solvefor x,0=x^2+2x-2m(x)^2 ?

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

Find the zeros of solvefor x,0=x^2+2x-2m(x)^2

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

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x, 0 = x^{2} + 2 x - 2 m (x)^{2}

x = 0, x = - \frac{2}{1 - 2 m}; m \neq = \frac{1}{2}

Solution steps

0 = x^{2} + 2 x - 2 m x^{2}

Switch sides

x^{2} + 2 x - 2 m x^{2} = 0

Write in the standard form

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

(1 - 2 m) x^{2} + 2 x = 0

Solve with the quadratic formula

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

2^{2} - 4 (1 - 2 m) \cdot 0 = 2

x_{1, 2} = \frac{- 2 \pm 2}{2 ( 1 - 2 m )}; m \neq = \frac{1}{2}

Separate the solutions

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

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

x = \frac{- 2 - 2}{2 ( 1 - 2 m )} : - \frac{2}{1 - 2 m}

The solutions to the quadratic equation are:

x = 0, x = - \frac{2}{1 - 2 m}; m \neq = \frac{1}{2}

\frac{( x - 1 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{64} = 1

\frac{( x + 1 ) ^{2}}{3} - \frac{( y - 1 ) ^{2}}{5} = 1

- 36 x^{2} + y^{2} = 36

16 x^{2} - 9 y^{2} + 96 x + 36 y + 252 = 0

4 y^{2} - x^{2} = 36

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