Solve sqrt{x+9}=1+sqrt{x}

$$\sqrt{ x+9 } =1+ \sqrt{ x }$$

Solve for x

$$x=16$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+9}\right)^{2}=\left(1+\sqrt{x}\right)^{2}$$

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

$$x+9=\left(1+\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{x}\right)^{2}.

$$x+9=1+2\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x+9=1+2\sqrt{x}+x$$

Subtract 2\sqrt{x} from both sides.

$$x+9-2\sqrt{x}=1+x$$

Subtract x from both sides.

$$x+9-2\sqrt{x}-x=1$$

Combine x and -x to get 0.

$$9-2\sqrt{x}=1$$

Subtract 9 from both sides.

$$-2\sqrt{x}=1-9$$

Subtract 9 from 1 to get -8.

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

Divide both sides by -2.

$$\sqrt{x}=\frac{-8}{-2}$$

Divide -8 by -2 to get 4.

$$\sqrt{x}=4$$

Square both sides of the equation.

$$x=16$$

Substitute 16 for x in the equation \sqrt{x+9}=1+\sqrt{x}.

$$\sqrt{16+9}=1+\sqrt{16}$$

Simplify. The value x=16 satisfies the equation.

$$5=5$$

Equation \sqrt{x+9}=\sqrt{x}+1 has a unique solution.

$$x=16$$