Solve sqrt{x+6}-sqrt{x-2}=2

$$\sqrt{ x+6 } - \sqrt{ x-2 } = 2$$

Solve for x

$$x=3$$

Solution Steps

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

$$\sqrt{x+6}=2+\sqrt{x-2}$$

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(2+\sqrt{x-2}\right)^{2}$$

Calculate \sqrt{x+6} to the power of 2 and get x+6.

$$x+6=\left(2+\sqrt{x-2}\right)^{2}$$

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

$$x+6=4+4\sqrt{x-2}+\left(\sqrt{x-2}\right)^{2}$$

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

$$x+6=4+4\sqrt{x-2}+x-2$$

Subtract 2 from 4 to get 2.

$$x+6=2+4\sqrt{x-2}+x$$

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

$$x+6-4\sqrt{x-2}=2+x$$

Subtract x from both sides.

$$x+6-4\sqrt{x-2}-x=2$$

Combine x and -x to get 0.

$$6-4\sqrt{x-2}=2$$

Subtract 6 from both sides.

$$-4\sqrt{x-2}=2-6$$

Subtract 6 from 2 to get -4.

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

Divide both sides by -4.

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

Divide -4 by -4 to get 1.

$$\sqrt{x-2}=1$$

Square both sides of the equation.

$$x-2=1$$

Add 2 to both sides of the equation.

$$x-2-\left(-2\right)=1-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x=1-\left(-2\right)$$

Subtract -2 from 1.

$$x=3$$

Substitute 3 for x in the equation \sqrt{x+6}-\sqrt{x-2}=2.

$$\sqrt{3+6}-\sqrt{3-2}=2$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Equation \sqrt{x+6}=\sqrt{x-2}+2 has a unique solution.

$$x=3$$