Solve f(x)=x^2+1quadg(x)=-4x-1

$$f ( x ) = x ^ { 2 } + 1 \quad g ( x ) = - 4 x - 1$$

Solve for g

$$g=-x-4-\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$1gx=-4x-1-x^{2}$$

Reorder the terms.

$$gx=-x^{2}-4x-1$$

The equation is in standard form.

$$xg=-x^{2}-4x-1$$

Divide both sides by x.

$$\frac{xg}{x}=\frac{-x^{2}-4x-1}{x}$$

Dividing by x undoes the multiplication by x.

$$g=\frac{-x^{2}-4x-1}{x}$$

Divide -x^{2}-4x-1 by x.

$$g=-x-4-\frac{1}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

$$x=\frac{-\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

$$x=\frac{-\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}\text{, }g\leq -6\text{ or }g\geq -2$$