Solve y=sqrt{3x-12}

$$y = \sqrt { 3 x - 12 }$$

Solve for x

$$x=\frac{y^{2}+12}{3}$$

$$y\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{3x-12}=y$$

Square both sides of the equation.

$$3x-12=y^{2}$$

Add 12 to both sides of the equation.

$$3x-12-\left(-12\right)=y^{2}-\left(-12\right)$$

Subtracting -12 from itself leaves 0.

$$3x=y^{2}-\left(-12\right)$$

Subtract -12 from y^{2}.

$$3x=y^{2}+12$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{y^{2}+12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{y^{2}+12}{3}$$

Divide y^{2}+12 by 3.

$$x=\frac{y^{2}}{3}+4$$

Solve for x (complex solution)

$$x=\frac{y^{2}+12}{3}$$

$$arg(y)<\pi \text{ or }y=0$$

Solve for y (complex solution)

$$y=\sqrt{3\left(x-4\right)}$$

Solve for y

$$y=\sqrt{3\left(x-4\right)}$$

$$x\geq 4$$