Solve sqrt{x-3}+sqrt{x}=3

$$\sqrt { x - 3 } + \sqrt { x } = 3$$

Solve for x

$$x=4$$

Solution Steps

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

$$\sqrt{x-3}=3-\sqrt{x}$$

Square both sides of the equation.

$$\left(\sqrt{x-3}\right)^{2}=\left(3-\sqrt{x}\right)^{2}$$

Calculate \sqrt{x-3} to the power of 2 and get x-3.

$$x-3=\left(3-\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{x}\right)^{2}.

$$x-3=9-6\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x-3=9-6\sqrt{x}+x$$

Add 6\sqrt{x} to both sides.

$$x-3+6\sqrt{x}=9+x$$

Subtract x from both sides.

$$x-3+6\sqrt{x}-x=9$$

Combine x and -x to get 0.

$$-3+6\sqrt{x}=9$$

Add 3 to both sides.

$$6\sqrt{x}=9+3$$

Add 9 and 3 to get 12.

$$6\sqrt{x}=12$$

Divide both sides by 6.

$$\sqrt{x}=\frac{12}{6}$$

Divide 12 by 6 to get 2.

$$\sqrt{x}=2$$

Square both sides of the equation.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x-3}+\sqrt{x}=3.

$$\sqrt{4-3}+\sqrt{4}=3$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation \sqrt{x-3}=-\sqrt{x}+3 has a unique solution.

$$x=4$$