Solve p^2x+px+4=p^2+x

$$p ^ { 2 } x + p x + 4 = p ^ { 2 } + x$$

Solve for x

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

$$p\neq \frac{\sqrt{5}-1}{2}\text{ and }p\neq \frac{-\sqrt{5}-1}{2}$$

Steps for Solving Linear Equation

Subtract x from both sides.

$$p^{2}x+px+4-x=p^{2}$$

Subtract 4 from both sides.

$$p^{2}x+px-x=p^{2}-4$$

Combine all terms containing x.

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

Divide both sides by p^{2}+p-1.

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

Dividing by p^{2}+p-1 undoes the multiplication by p^{2}+p-1.

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

Solve for p (complex solution)

$$\left\{\begin{matrix}p=\frac{\sqrt{5x^{2}-20x+16}-x}{2\left(x-1\right)}\text{; }p=-\frac{\sqrt{5x^{2}-20x+16}+x}{2\left(x-1\right)}\text{, }&x\neq 1\\p=-3\text{, }&x=1\end{matrix}\right.$$

Solve for p

$$\left\{\begin{matrix}p=\frac{\sqrt{5x^{2}-20x+16}-x}{2\left(x-1\right)}\text{; }p=-\frac{\sqrt{5x^{2}-20x+16}+x}{2\left(x-1\right)}\text{, }&\left(x\neq 1\text{ and }x\leq -\frac{2\sqrt{5}}{5}+2\right)\text{ or }x\geq \frac{2\sqrt{5}}{5}+2\\p=-3\text{, }&x=1\end{matrix}\right.$$