Solve y=sqrt{ax+b}

$$y = \sqrt { a x + b }$$

Solve for a

$$\left\{\begin{matrix}a\in \mathrm{R}\text{, }&x=0\text{ and }y\geq 0\text{ and }b=y^{2}\\a=-\frac{b-y^{2}}{x}\text{, }&y\geq 0\text{ and }x\neq 0\end{matrix}\right.$$

Steps for Solving Radical Equations

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

$$\sqrt{ax+b}=y$$

Square both sides of the equation.

$$xa+b=y^{2}$$

Subtract b from both sides of the equation.

$$xa+b-b=y^{2}-b$$

Subtracting b from itself leaves 0.

$$xa=y^{2}-b$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{y^{2}-b}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{y^{2}-b}{x}$$

Solve for b

$$b=y^{2}-ax$$

$$y\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{ax+b}=y$$

Square both sides of the equation.

$$b+ax=y^{2}$$

Subtract ax from both sides of the equation.

$$b+ax-ax=y^{2}-ax$$

Subtracting ax from itself leaves 0.

$$b=y^{2}-ax$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{b-y^{2}}{x}\text{, }&x\neq 0\text{ and }\left(y=0\text{ or }arg(y)<\pi \right)\\a\in \mathrm{C}\text{, }&\left(b=0\text{ and }x=0\text{ and }y=0\right)\text{ or }\left(b=y^{2}\text{ and }x=0\text{ and }arg(y)<\pi \text{ and }y\neq 0\right)\end{matrix}\right.$$

Solve for b (complex solution)

$$b=y^{2}-ax$$

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