Solve sqrt{x+8}=1+sqrt{x+1}

$$\sqrt { x + 8 } = 1 + \sqrt { x + 1 }$$

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+8}\right)^{2}=\left(1+\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+8} to the power of 2 and get x+8.

$$x+8=\left(1+\sqrt{x+1}\right)^{2}$$

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

$$x+8=1+2\sqrt{x+1}+\left(\sqrt{x+1}\right)^{2}$$

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

$$x+8=1+2\sqrt{x+1}+x+1$$

Add 1 and 1 to get 2.

$$x+8=2+2\sqrt{x+1}+x$$

Subtract 2\sqrt{x+1} from both sides.

$$x+8-2\sqrt{x+1}=2+x$$

Subtract x from both sides.

$$x+8-2\sqrt{x+1}-x=2$$

Combine x and -x to get 0.

$$8-2\sqrt{x+1}=2$$

Subtract 8 from both sides.

$$-2\sqrt{x+1}=2-8$$

Subtract 8 from 2 to get -6.

$$-2\sqrt{x+1}=-6$$

Divide both sides by -2.

$$\sqrt{x+1}=\frac{-6}{-2}$$

Divide -6 by -2 to get 3.

$$\sqrt{x+1}=3$$

Square both sides of the equation.

$$x+1=9$$

Subtract 1 from both sides of the equation.

$$x+1-1=9-1$$

Subtracting 1 from itself leaves 0.

$$x=9-1$$

Subtract 1 from 9.

$$x=8$$

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

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

Simplify. The value x=8 satisfies the equation.

$$4=4$$

Equation \sqrt{x+8}=\sqrt{x+1}+1 has a unique solution.

$$x=8$$