Solve sqrt{x+5}+sqrt{x}=5

$$\sqrt { x + 5 } + \sqrt { x } = 5$$

Solve for x

$$x=4$$

Solution Steps

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

$$\sqrt{x+5}=5-\sqrt{x}$$

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(5-\sqrt{x}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x+5=\left(5-\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x}\right)^{2}.

$$x+5=25-10\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x+5=25-10\sqrt{x}+x$$

Add 10\sqrt{x} to both sides.

$$x+5+10\sqrt{x}=25+x$$

Subtract x from both sides.

$$x+5+10\sqrt{x}-x=25$$

Combine x and -x to get 0.

$$5+10\sqrt{x}=25$$

Subtract 5 from both sides.

$$10\sqrt{x}=25-5$$

Subtract 5 from 25 to get 20.

$$10\sqrt{x}=20$$

Divide both sides by 10.

$$\sqrt{x}=\frac{20}{10}$$

Divide 20 by 10 to get 2.

$$\sqrt{x}=2$$

Square both sides of the equation.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x+5}+\sqrt{x}=5.

$$\sqrt{4+5}+\sqrt{4}=5$$

Simplify. The value x=4 satisfies the equation.

$$5=5$$

Equation \sqrt{x+5}=-\sqrt{x}+5 has a unique solution.

$$x=4$$