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

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

Solve for x

$$x = \frac{25}{4} = 6\frac{1}{4} = 6.25$$

Solution Steps

Subtract \sqrt{x-4} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$x=16-8\sqrt{x-4}+\left(\sqrt{x-4}\right)^{2}$$

Calculate \sqrt{x-4} to the power of 2 and get x-4.

$$x=16-8\sqrt{x-4}+x-4$$

Subtract 4 from 16 to get 12.

$$x=12-8\sqrt{x-4}+x$$

Add 8\sqrt{x-4} to both sides.

$$x+8\sqrt{x-4}=12+x$$

Subtract x from both sides.

$$x+8\sqrt{x-4}-x=12$$

Combine x and -x to get 0.

$$8\sqrt{x-4}=12$$

Divide both sides by 8.

$$\sqrt{x-4}=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$\sqrt{x-4}=\frac{3}{2}$$

Square both sides of the equation.

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

Add 4 to both sides of the equation.

$$x-4-\left(-4\right)=\frac{9}{4}-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$x=\frac{9}{4}-\left(-4\right)$$

Subtract -4 from \frac{9}{4}.

$$x=\frac{25}{4}$$

Substitute \frac{25}{4} for x in the equation \sqrt{x}+\sqrt{x-4}=4.

$$\sqrt{\frac{25}{4}}+\sqrt{\frac{25}{4}-4}=4$$

Simplify. The value x=\frac{25}{4} satisfies the equation.

$$4=4$$

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

$$x=\frac{25}{4}$$