Solve sqrt{y+a}=b

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

Solve for a

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

$$b\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

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

Subtract y from both sides of the equation.

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

Subtracting y from itself leaves 0.

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

Solve for a (complex solution)

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

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

Solve for b (complex solution)

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

Solve for b

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

$$y\geq -a$$