Solve q=sqrt{L}+sqrt{8}

$$q = \sqrt { L } + \sqrt { 8 }$$

Solve for L

$$L=\left(q-2\sqrt{2}\right)^{2}$$

$$q-2\sqrt{2}\geq 0$$

Steps for Solving Radical Equations

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$q=\sqrt{L}+2\sqrt{2}$$

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

$$\sqrt{L}+2\sqrt{2}=q$$

Subtract 2\sqrt{2} from both sides.

$$\sqrt{L}=q-2\sqrt{2}$$

Square both sides of the equation.

$$L=\left(q-2\sqrt{2}\right)^{2}$$

Solve for L (complex solution)

$$L=\left(q-2\sqrt{2}\right)^{2}$$

$$q=2\sqrt{2}\text{ or }arg(q-2\sqrt{2})<\pi$$

Solve for q (complex solution)

$$q=\sqrt{L}+2\sqrt{2}$$

Solve for q

$$q=\sqrt{L}+2\sqrt{2}$$

$$L\geq 0$$