Solve sqrt{x+9}=9-sqrt{x}

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

Solve for x

$$x=16$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x+9} to the power of 2 and get x+9.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-\sqrt{x}\right)^{2}.

$$x+9=81-18\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x+9=81-18\sqrt{x}+x$$

Add 18\sqrt{x} to both sides.

$$x+9+18\sqrt{x}=81+x$$

Subtract x from both sides.

$$x+9+18\sqrt{x}-x=81$$

Combine x and -x to get 0.

$$9+18\sqrt{x}=81$$

Subtract 9 from both sides.

$$18\sqrt{x}=81-9$$

Subtract 9 from 81 to get 72.

$$18\sqrt{x}=72$$

Divide both sides by 18.

$$\sqrt{x}=\frac{72}{18}$$

Divide 72 by 18 to get 4.

$$\sqrt{x}=4$$

Square both sides of the equation.

$$x=16$$

Substitute 16 for x in the equation \sqrt{x+9}=9-\sqrt{x}.

$$\sqrt{16+9}=9-\sqrt{16}$$

Simplify. The value x=16 satisfies the equation.

$$5=5$$

Equation \sqrt{x+9}=-\sqrt{x}+9 has a unique solution.

$$x=16$$