Solve sqrt{x+6}=1+sqrt{x+1}

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(1+\sqrt{x+1}\right)^{2}$$

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

$$x+6=\left(1+\sqrt{x+1}\right)^{2}$$

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

$$x+6=1+2\sqrt{x+1}+\left(\sqrt{x+1}\right)^{2}$$

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

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

Add 1 and 1 to get 2.

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

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

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Subtract 6 from both sides.

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

Subtract 6 from 2 to get -4.

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

Divide both sides by -2.

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

Divide -4 by -2 to get 2.

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

Square both sides of the equation.

$$x+1=4$$

Subtract 1 from both sides of the equation.

$$x+1-1=4-1$$

Subtracting 1 from itself leaves 0.

$$x=4-1$$

Subtract 1 from 4.

$$x=3$$

Substitute 3 for x in the equation \sqrt{x+6}=1+\sqrt{x+1}.

$$\sqrt{3+6}=1+\sqrt{3+1}$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Equation \sqrt{x+6}=\sqrt{x+1}+1 has a unique solution.

$$x=3$$