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

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

Solve for x

$$x=2$$

Solution Steps

Subtract -\sqrt{x-2} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 2 from 4 to get 2.

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

Subtract 4\sqrt{x-2} from both sides.

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

$$-4\sqrt{x-2}=0$$

Divide both sides by -4. Zero divided by any non-zero number gives zero.

$$\sqrt{x-2}=0$$

Square both sides of the equation.

$$x-2=0$$

Add 2 to both sides of the equation.

$$x-2-\left(-2\right)=-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x=-\left(-2\right)$$

Subtract -2 from 0.

$$x=2$$

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

$$\sqrt{2+2}-\sqrt{2-2}=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

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

$$x=2$$