Solve f(x)=1/x^2+2xquadg(x)=1-2x/x

$$f ( x ) = \frac { 1 } { x ^ { 2 } + 2 x } \quad g ( x ) = \frac { 1 - 2 x } { x }$$

Solve for g

$$g=-2x-3+\frac{2}{x}$$

$$x\neq -2\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x\left(x+2\right), the least common multiple of x^{2}+2x,x.

$$1gx=\left(x+2\right)\left(1-2x\right)$$

Use the distributive property to multiply x+2 by 1-2x and combine like terms.

$$1gx=-3x-2x^{2}+2$$

Reorder the terms.

$$gx=-2x^{2}-3x+2$$

The equation is in standard form.

$$xg=2-3x-2x^{2}$$

Divide both sides by x.

$$\frac{xg}{x}=-\frac{\left(2x-1\right)\left(x+2\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$g=-\frac{\left(2x-1\right)\left(x+2\right)}{x}$$

Divide -\left(2+x\right)\left(-1+2x\right) by x.

$$g=-2x-3+\frac{2}{x}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{\sqrt{g^{2}+6g+25}-g-3}{4}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{g^{2}+6g+25}-g-3}{4}\text{, }&g\neq 0\end{matrix}\right.$$