Solve sqrt{8X}+9=x

$$\sqrt { 8 X } + 9 = x$$

Solve for X

$$X=\frac{\left(9-x\right)^{2}}{8}$$

$$-\left(9-x\right)\geq 0$$

Steps for Solving Radical Equations

Subtract 9 from both sides of the equation.

$$\sqrt{8X}+9-9=x-9$$

Subtracting 9 from itself leaves 0.

$$\sqrt{8X}=x-9$$

Square both sides of the equation.

$$8X=\left(x-9\right)^{2}$$

Divide both sides by 8.

$$\frac{8X}{8}=\frac{\left(x-9\right)^{2}}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$X=\frac{\left(x-9\right)^{2}}{8}$$

Solve for X (complex solution)

$$X=\frac{\left(9-x\right)^{2}}{8}$$

$$x=9\text{ or }arg(9-x)\geq \pi$$

Solve for x (complex solution)

$$x=2\sqrt{2X}+9$$

Solve for x

$$x=2\sqrt{2X}+9$$

$$X\geq 0$$