Solve x-sqrt{x2-4x}=3

$$x - \sqrt { x 2 - 4 x } = 3$$

Solve for x_2

$$x_{2}=\left(x-3\right)^{2}+4x$$

$$x-3\geq 0$$

Steps for Solving Radical Equations

Subtract x from both sides of the equation.

$$-\sqrt{x_{2}-4x}+x-x=3-x$$

Subtracting x from itself leaves 0.

$$-\sqrt{x_{2}-4x}=3-x$$

Divide both sides by -1.

$$\frac{-\sqrt{x_{2}-4x}}{-1}=\frac{3-x}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\sqrt{x_{2}-4x}=\frac{3-x}{-1}$$

Divide 3-x by -1.

$$\sqrt{x_{2}-4x}=x-3$$

Square both sides of the equation.

$$x_{2}-4x=\left(x-3\right)^{2}$$

Subtract -4x from both sides of the equation.

$$x_{2}-4x-\left(-4x\right)=\left(x-3\right)^{2}-\left(-4x\right)$$

Subtracting -4x from itself leaves 0.

$$x_{2}=\left(x-3\right)^{2}-\left(-4x\right)$$

Subtract -4x from \left(x-3\right)^{2}.

$$x_{2}=\left(x-3\right)^{2}+4x$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\sqrt{x_{2}-8}+1\text{, }&x_{2}=12\text{ or }arg(\sqrt{x_{2}-8}-2)<\pi \\x=-\sqrt{x_{2}-8}+1\text{, }&arg(-\sqrt{x_{2}-8}-2)<\pi \end{matrix}\right.$$

Solve for x_2 (complex solution)

$$x_{2}=\left(x-3\right)^{2}+4x$$

$$x=3\text{ or }arg(x-3)<\pi$$

Solve for x

$$x=\sqrt{x_{2}-8}+1$$

$$x_{2}\geq 12$$