Solve sqrt{x-4}+2=sqrt{x+12}

$$\sqrt{ x-4 } +2 = \sqrt{ x+12 }$$

Solve for x

$$x=13$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x-4}+2\right)^{2}=\left(\sqrt{x+12}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x-4}+2\right)^{2}.

$$\left(\sqrt{x-4}\right)^{2}+4\sqrt{x-4}+4=\left(\sqrt{x+12}\right)^{2}$$

Calculate \sqrt{x-4} to the power of 2 and get x-4.

$$x-4+4\sqrt{x-4}+4=\left(\sqrt{x+12}\right)^{2}$$

Add -4 and 4 to get 0.

$$x+4\sqrt{x-4}=\left(\sqrt{x+12}\right)^{2}$$

Calculate \sqrt{x+12} to the power of 2 and get x+12.

$$x+4\sqrt{x-4}=x+12$$

Subtract x from both sides.

$$x+4\sqrt{x-4}-x=12$$

Combine x and -x to get 0.

$$4\sqrt{x-4}=12$$

Divide both sides by 4.

$$\sqrt{x-4}=\frac{12}{4}$$

Divide 12 by 4 to get 3.

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

Square both sides of the equation.

$$x-4=9$$

Add 4 to both sides of the equation.

$$x-4-\left(-4\right)=9-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$x=9-\left(-4\right)$$

Subtract -4 from 9.

$$x=13$$

Substitute 13 for x in the equation \sqrt{x-4}+2=\sqrt{x+12}.

$$\sqrt{13-4}+2=\sqrt{13+12}$$

Simplify. The value x=13 satisfies the equation.

$$5=5$$

Equation \sqrt{x-4}+2=\sqrt{x+12} has a unique solution.

$$x=13$$