Solve sqrt{x+3}-sqrt{x-5}=2

$$\sqrt { x + 3 } - \sqrt { x - 5 } = 2$$

Solve for x

$$x=6$$

Solution Steps

Subtract -\sqrt{x-5} from both sides of the equation.

$$\sqrt{x+3}=2+\sqrt{x-5}$$

Square both sides of the equation.

$$\left(\sqrt{x+3}\right)^{2}=\left(2+\sqrt{x-5}\right)^{2}$$

Calculate \sqrt{x+3} to the power of 2 and get x+3.

$$x+3=\left(2+\sqrt{x-5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{x-5}\right)^{2}.

$$x+3=4+4\sqrt{x-5}+\left(\sqrt{x-5}\right)^{2}$$

Calculate \sqrt{x-5} to the power of 2 and get x-5.

$$x+3=4+4\sqrt{x-5}+x-5$$

Subtract 5 from 4 to get -1.

$$x+3=-1+4\sqrt{x-5}+x$$

Subtract 4\sqrt{x-5} from both sides.

$$x+3-4\sqrt{x-5}=-1+x$$

Subtract x from both sides.

$$x+3-4\sqrt{x-5}-x=-1$$

Combine x and -x to get 0.

$$3-4\sqrt{x-5}=-1$$

Subtract 3 from both sides.

$$-4\sqrt{x-5}=-1-3$$

Subtract 3 from -1 to get -4.

$$-4\sqrt{x-5}=-4$$

Divide both sides by -4.

$$\sqrt{x-5}=\frac{-4}{-4}$$

Divide -4 by -4 to get 1.

$$\sqrt{x-5}=1$$

Square both sides of the equation.

$$x-5=1$$

Add 5 to both sides of the equation.

$$x-5-\left(-5\right)=1-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$x=1-\left(-5\right)$$

Subtract -5 from 1.

$$x=6$$

Substitute 6 for x in the equation \sqrt{x+3}-\sqrt{x-5}=2.

$$\sqrt{6+3}-\sqrt{6-5}=2$$

Simplify. The value x=6 satisfies the equation.

$$2=2$$

Equation \sqrt{x+3}=\sqrt{x-5}+2 has a unique solution.

$$x=6$$