Solve x^2+alphax+4alpha+1=0

$$x ^ { 2 } + \alpha x + 4 \alpha + 1 = 0$$

Solve for α

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

$$x\neq -4$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$\alpha x+4\alpha +1=-x^{2}$$

Subtract 1 from both sides.

$$\alpha x+4\alpha =-x^{2}-1$$

Combine all terms containing \alpha .

$$\left(x+4\right)\alpha =-x^{2}-1$$

Divide both sides by x+4.

$$\frac{\left(x+4\right)\alpha }{x+4}=\frac{-x^{2}-1}{x+4}$$

Dividing by x+4 undoes the multiplication by x+4.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\alpha ^{2}-16\alpha -4}-\alpha }{2}$$

$$x=\frac{-\sqrt{\alpha ^{2}-16\alpha -4}-\alpha }{2}$$

Solve for x

$$x=\frac{\sqrt{\alpha ^{2}-16\alpha -4}-\alpha }{2}$$

$$x=\frac{-\sqrt{\alpha ^{2}-16\alpha -4}-\alpha }{2}\text{, }\alpha \geq 2\sqrt{17}+8\text{ or }\alpha \leq 8-2\sqrt{17}$$