Solve sqrt{x}-sqrt{x+6}+sqrt{2}=0

$$\sqrt { x } - \sqrt { x + 6 } + \sqrt { 2 } = 0$$

Solve for x

$$x=2$$

Solution Steps

Subtract -\sqrt{x+6}+\sqrt{2} from both sides of the equation.

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

To find the opposite of -\sqrt{x+6}+\sqrt{2}, find the opposite of each term.

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

The opposite of -\sqrt{x+6} is \sqrt{x+6}.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{x+6}-\sqrt{2}\right)^{2}.

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

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

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

$$x=x+8-2\sqrt{x+6}\sqrt{2}$$

Subtract x from both sides.

$$x-x=8-2\sqrt{x+6}\sqrt{2}$$

Combine x and -x to get 0.

$$0=8-2\sqrt{x+6}\sqrt{2}$$

Swap sides so that all variable terms are on the left hand side.

$$8-2\sqrt{x+6}\sqrt{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-2\sqrt{x+6}\sqrt{2}=-8$$

Divide both sides by -2\sqrt{2}.

$$\frac{\left(-2\sqrt{2}\right)\sqrt{x+6}}{-2\sqrt{2}}=-\frac{8}{-2\sqrt{2}}$$

Dividing by -2\sqrt{2} undoes the multiplication by -2\sqrt{2}.

$$\sqrt{x+6}=-\frac{8}{-2\sqrt{2}}$$

Divide -8 by -2\sqrt{2}.

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

Square both sides of the equation.

$$x+6=8$$

Subtract 6 from both sides of the equation.

$$x+6-6=8-6$$

Subtracting 6 from itself leaves 0.

$$x=8-6$$

Subtract 6 from 8.

$$x=2$$

Substitute 2 for x in the equation \sqrt{x}-\sqrt{x+6}+\sqrt{2}=0.

$$\sqrt{2}-\sqrt{2+6}+\sqrt{2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

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

$$x=2$$