Solve sqrt{a^2+3b}+a=8

$$\sqrt { a ^ { 2 } + 3 b } + a = 8$$

Solve for b

$$b=\frac{64-16a}{3}$$

$$8-a\geq 0$$

Steps for Solving Radical Equations

Subtract a from both sides of the equation.

$$\sqrt{3b+a^{2}}+a-a=8-a$$

Subtracting a from itself leaves 0.

$$\sqrt{3b+a^{2}}=8-a$$

Square both sides of the equation.

$$3b+a^{2}=\left(8-a\right)^{2}$$

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

$$3b+a^{2}-a^{2}=\left(8-a\right)^{2}-a^{2}$$

Subtracting a^{2} from itself leaves 0.

$$3b=\left(8-a\right)^{2}-a^{2}$$

Subtract a^{2} from \left(-a+8\right)^{2}.

$$3b=64-16a$$

Divide both sides by 3.

$$\frac{3b}{3}=\frac{64-16a}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$b=\frac{64-16a}{3}$$

Solve for a (complex solution)

$$a=-\frac{3b}{16}+4$$

$$b=-\frac{64}{3}\text{ or }arg(\frac{3b}{16}+4)<\pi$$

Solve for b (complex solution)

$$b=\frac{64-16a}{3}$$

$$a=8\text{ or }arg(8-a)<\pi$$

Solve for a

$$a=-\frac{3b}{16}+4$$

$$b\geq -\frac{64}{3}$$