Solve z=msqrt{y^2-2x}

$$z = m \sqrt { y ^ { 2 } - 2 x }$$

Solve for m

$$\left\{\begin{matrix}m=\frac{z}{\sqrt{y^{2}-2x}}\text{, }&x<\frac{y^{2}}{2}\\m\in \mathrm{R}\text{, }&x=\frac{y^{2}}{2}\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$m\sqrt{y^{2}-2x}=z$$

The equation is in standard form.

$$\sqrt{y^{2}-2x}m=z$$

Divide both sides by \sqrt{y^{2}-2x}.

$$\frac{\sqrt{y^{2}-2x}m}{\sqrt{y^{2}-2x}}=\frac{z}{\sqrt{y^{2}-2x}}$$

Dividing by \sqrt{y^{2}-2x} undoes the multiplication by \sqrt{y^{2}-2x}.

$$m=\frac{z}{\sqrt{y^{2}-2x}}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{y^{2}-\left(\frac{z}{m}\right)^{2}}{2}\text{, }&\left(z\geq 0\text{ and }m>0\right)\text{ or }\left(z\leq 0\text{ and }m<0\right)\\x\leq \frac{y^{2}}{2}\text{, }&z=0\text{ and }m=0\end{matrix}\right.$$

Steps for Solving Radical Equations

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

$$m\sqrt{y^{2}-2x}=z$$

Divide both sides by m.

$$\frac{m\sqrt{-2x+y^{2}}}{m}=\frac{z}{m}$$

Dividing by m undoes the multiplication by m.

$$\sqrt{-2x+y^{2}}=\frac{z}{m}$$

Square both sides of the equation.

$$-2x+y^{2}=\frac{z^{2}}{m^{2}}$$

Subtract y^{2} from both sides of the equation.

$$-2x+y^{2}-y^{2}=\frac{z^{2}}{m^{2}}-y^{2}$$

Subtracting y^{2} from itself leaves 0.

$$-2x=\frac{z^{2}}{m^{2}}-y^{2}$$

Subtract y^{2} from \frac{z^{2}}{m^{2}}.

$$-2x=-y^{2}+\frac{z^{2}}{m^{2}}$$

Divide both sides by -2.

$$\frac{-2x}{-2}=\frac{-y^{2}+\frac{z^{2}}{m^{2}}}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x=\frac{-y^{2}+\frac{z^{2}}{m^{2}}}{-2}$$

Divide -y^{2}+\frac{z^{2}}{m^{2}} by -2.

$$x=\frac{y^{2}}{2}-\frac{z^{2}}{2m^{2}}$$