Solve x+sqrt{xy}=x^2

$$x + \sqrt { x y } = x ^ { 2 }$$

Solve for y

$$\left\{\begin{matrix}y=x\left(x-1\right)^{2}\text{, }&x\leq 0\text{ or }x\geq 1\\y\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Radical Equations

Subtract x from both sides of the equation.

$$\sqrt{xy}+x-x=x^{2}-x$$

Subtracting x from itself leaves 0.

$$\sqrt{xy}=x^{2}-x$$

Subtract x from x^{2}.

$$\sqrt{xy}=x\left(x-1\right)$$

Square both sides of the equation.

$$xy=x^{2}\left(x-1\right)^{2}$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{x^{2}\left(x-1\right)^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{x^{2}\left(x-1\right)^{2}}{x}$$

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

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