Solve sqrt{x^4}sqrt{y}=x

$$\sqrt { x ^ { 4 } } \sqrt { y } = x$$

Solve for y

$$\left\{\begin{matrix}y=\frac{1}{x^{2}}\text{, }&x>0\\y\geq 0\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Radical Equations

Divide both sides by \sqrt{x^{4}}.

$$\frac{\sqrt{x^{4}}\sqrt{y}}{\sqrt{x^{4}}}=\frac{x}{\sqrt{x^{4}}}$$

Dividing by \sqrt{x^{4}} undoes the multiplication by \sqrt{x^{4}}.

$$\sqrt{y}=\frac{x}{\sqrt{x^{4}}}$$

Divide x by \sqrt{x^{4}}.

$$\sqrt{y}=\frac{1}{x}$$

Square both sides of the equation.

$$y=\frac{1}{x^{2}}$$

Solve for x

$$\left\{\begin{matrix}x=0\text{, }&y\geq 0\\x=\frac{1}{\sqrt{y}}\text{, }&y>0\end{matrix}\right.$$