Solve y=x[p+sqrt{(1+p^2)]}

$$y = x [ p + \sqrt { ( 1 + p ^ { 2 } ) ] }$$

Solve for x

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

Steps for Solving Linear Equation

Use the distributive property to multiply x by p+\sqrt{1+p^{2}}.

$$y=xp+x\sqrt{1+p^{2}}$$

Swap sides so that all variable terms are on the left hand side.

$$xp+x\sqrt{1+p^{2}}=y$$

Combine all terms containing x.

$$\left(p+\sqrt{1+p^{2}}\right)x=y$$

The equation is in standard form.

$$\left(\sqrt{p^{2}+1}+p\right)x=y$$

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

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

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

$$x=\frac{y}{\sqrt{p^{2}+1}+p}$$

Divide y by p+\sqrt{1+p^{2}}.

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

Solve for p

$$\left\{\begin{matrix}p=\frac{y^{2}-x^{2}}{2xy}\text{, }&\left(y<0\text{ and }x<0\right)\text{ or }\left(y>0\text{ and }x>0\right)\\p\in \mathrm{R}\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$