Solve sqrt{x-2}+sqrt{x}=4

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

Solve for x

$$x = \frac{81}{16} = 5\frac{1}{16} = 5.0625$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

$$x-2=16-8\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x-2=16-8\sqrt{x}+x$$

Add 8\sqrt{x} to both sides.

$$x-2+8\sqrt{x}=16+x$$

Subtract x from both sides.

$$x-2+8\sqrt{x}-x=16$$

Combine x and -x to get 0.

$$-2+8\sqrt{x}=16$$

Add 2 to both sides.

$$8\sqrt{x}=16+2$$

Add 16 and 2 to get 18.

$$8\sqrt{x}=18$$

Divide both sides by 8.

$$\sqrt{x}=\frac{18}{8}$$

Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.

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

Square both sides of the equation.

$$x=\frac{81}{16}$$

Substitute \frac{81}{16} for x in the equation \sqrt{x-2}+\sqrt{x}=4.

$$\sqrt{\frac{81}{16}-2}+\sqrt{\frac{81}{16}}=4$$

Simplify. The value x=\frac{81}{16} satisfies the equation.

$$4=4$$

Equation \sqrt{x-2}=-\sqrt{x}+4 has a unique solution.

$$x=\frac{81}{16}$$