Solve v=frac{sqrt{2x+1}}{5}

$$v = \frac { \sqrt { 2 x + 1 } } { 5 }$$

Solve for x

$$x=\frac{25v^{2}-1}{2}$$

$$v\geq 0$$

Steps for Solving Radical Equations

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

$$\frac{\sqrt{2x+1}}{5}=v$$

Multiply both sides by 5.

$$\frac{\frac{1}{5}\sqrt{2x+1}}{\frac{1}{5}}=\frac{v}{\frac{1}{5}}$$

Dividing by \frac{1}{5} undoes the multiplication by \frac{1}{5}.

$$\sqrt{2x+1}=\frac{v}{\frac{1}{5}}$$

Divide v by \frac{1}{5} by multiplying v by the reciprocal of \frac{1}{5}.

$$\sqrt{2x+1}=5v$$

Square both sides of the equation.

$$2x+1=25v^{2}$$

Subtract 1 from both sides of the equation.

$$2x+1-1=25v^{2}-1$$

Subtracting 1 from itself leaves 0.

$$2x=25v^{2}-1$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{25v^{2}-1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{25v^{2}-1}{2}$$

Solve for x (complex solution)

$$x=\frac{25v^{2}-1}{2}$$

$$arg(v)<\pi \text{ or }v=0$$

Solve for v

$$v=\frac{\sqrt{2x+1}}{5}$$

$$x\geq -\frac{1}{2}$$