Solve 2y=4sqrt{x^2-8}

$$2 y = 4 \sqrt { x ^ { 2 } - 8 }$$

Solve for y (complex solution)

$$y=2\sqrt{x^{2}-8}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$2y=4\sqrt{x^{2}-8}$$

Divide both sides by 2.

$$\frac{2y}{2}=\frac{4\sqrt{x^{2}-8}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y=\frac{4\sqrt{x^{2}-8}}{2}$$

Divide 4\sqrt{x^{2}-8} by 2.

$$y=2\sqrt{x^{2}-8}$$

Solve for y

$$y=2\sqrt{x^{2}-8}$$

$$|x|\geq 2\sqrt{2}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$2y=4\sqrt{x^{2}-8}$$

Divide both sides by 2.

$$\frac{2y}{2}=\frac{4\sqrt{x^{2}-8}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y=\frac{4\sqrt{x^{2}-8}}{2}$$

Divide 4\sqrt{x^{2}-8} by 2.

$$y=2\sqrt{x^{2}-8}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{y^{2}+32}}{2}$$

$$x=\frac{\sqrt{y^{2}+32}}{2}\text{, }arg(y)<\pi \text{ or }y=0$$

Solve for x

$$x=\frac{\sqrt{y^{2}+32}}{2}$$

$$x=-\frac{\sqrt{y^{2}+32}}{2}\text{, }y\geq 0$$