Solve v=sqrt{v^2x+v^2}

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

Solve for v (complex solution)

$$\left\{\begin{matrix}\\v=0\text{, }&\text{unconditionally}\\v\in \mathrm{C}\text{, }&x=0\text{ and }arg(v)<\pi \text{ and }v\neq 0\end{matrix}\right.$$

Solution Steps

Square both sides of the equation.

$$v^{2}=\left(\sqrt{v^{2}x+v^{2}}\right)^{2}$$

Calculate \sqrt{v^{2}x+v^{2}} to the power of 2 and get v^{2}x+v^{2}.

$$v^{2}=v^{2}x+v^{2}$$

Subtract v^{2}x from both sides.

$$v^{2}-v^{2}x=v^{2}$$

Subtract v^{2} from both sides.

$$v^{2}-v^{2}x-v^{2}=0$$

Combine v^{2} and -v^{2} to get 0.

$$-v^{2}x=0$$

Dividing by -x undoes the multiplication by -x.

$$v^{2}=\frac{0}{-x}$$

Divide 0 by -x.

$$v^{2}=0$$

Take the square root of both sides of the equation.

$$v=0 v=0$$

The equation is now solved. Solutions are the same.

$$v=0$$

Substitute 0 for v in the equation v=\sqrt{v^{2}x+v^{2}}.

$$0=\sqrt{0^{2}x+0^{2}}$$

Simplify. The value v=0 satisfies the equation.

$$0=0$$

Equation v=\sqrt{xv^{2}+v^{2}} has a unique solution.

$$v=0$$

Solve for x

$$\left\{\begin{matrix}x=0\text{, }&v\geq 0\\x\geq -1\text{, }&v=0\end{matrix}\right.$$

Steps for Solving Radical Equations

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

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

Square both sides of the equation.

$$v^{2}x+v^{2}=v^{2}$$

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

$$v^{2}x+v^{2}-v^{2}=v^{2}-v^{2}$$

Subtracting v^{2} from itself leaves 0.

$$v^{2}x=v^{2}-v^{2}$$

Subtract v^{2} from v^{2}.

$$v^{2}x=0$$

Divide both sides by v^{2}.

$$\frac{v^{2}x}{v^{2}}=\frac{0}{v^{2}}$$

Dividing by v^{2} undoes the multiplication by v^{2}.

$$x=\frac{0}{v^{2}}$$

Divide 0 by v^{2}.

$$x=0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=0\text{, }&arg(v)<\pi \text{ or }v=0\\x\in \mathrm{C}\text{, }&v=0\end{matrix}\right.$$

Solve for v

$$\left\{\begin{matrix}v=0\text{, }&x\geq -1\\v\geq 0\text{, }&x=0\end{matrix}\right.$$