Solve sqrt{x+1}-sqrt{x-4}=1

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

Solve for x

$$x=8$$

Solution Steps

Subtract -\sqrt{x-4} from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{x+1}\right)^{2}=\left(1+\sqrt{x-4}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x+1=\left(1+\sqrt{x-4}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{x-4}\right)^{2}.

$$x+1=1+2\sqrt{x-4}+\left(\sqrt{x-4}\right)^{2}$$

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

$$x+1=1+2\sqrt{x-4}+x-4$$

Subtract 4 from 1 to get -3.

$$x+1=-3+2\sqrt{x-4}+x$$

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

$$x+1-2\sqrt{x-4}=-3+x$$

Subtract x from both sides.

$$x+1-2\sqrt{x-4}-x=-3$$

Combine x and -x to get 0.

$$1-2\sqrt{x-4}=-3$$

Subtract 1 from both sides.

$$-2\sqrt{x-4}=-3-1$$

Subtract 1 from -3 to get -4.

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

Divide both sides by -2.

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

Divide -4 by -2 to get 2.

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

Square both sides of the equation.

$$x-4=4$$

Add 4 to both sides of the equation.

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

Subtracting -4 from itself leaves 0.

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

Subtract -4 from 4.

$$x=8$$

Substitute 8 for x in the equation \sqrt{x+1}-\sqrt{x-4}=1.

$$\sqrt{8+1}-\sqrt{8-4}=1$$

Simplify. The value x=8 satisfies the equation.

$$1=1$$

Equation \sqrt{x+1}=\sqrt{x-4}+1 has a unique solution.

$$x=8$$