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

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

Solve for x

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 2 from 4 to get 2.

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

Add 4\sqrt{x-2} to both sides.

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Divide both sides by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Square both sides of the equation.

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

Add 2 to both sides of the equation.

$$x-2-\left(-2\right)=\frac{1}{4}-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x=\frac{1}{4}-\left(-2\right)$$

Subtract -2 from \frac{1}{4}.

$$x=\frac{9}{4}$$

Substitute \frac{9}{4} for x in the equation \sqrt{x}+\sqrt{x-2}=2.

$$\sqrt{\frac{9}{4}}+\sqrt{\frac{9}{4}-2}=2$$

Simplify. The value x=\frac{9}{4} satisfies the equation.

$$2=2$$

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

$$x=\frac{9}{4}$$