Solve sqrt{2x+33}-sqrt{2x}=3

$$\sqrt { 2 x + 33 } - \sqrt { 2 x } = 3$$

Solve for x

$$x=8$$

Solution Steps

Subtract -\sqrt{2x} from both sides of the equation.

$$\sqrt{2x+33}=3+\sqrt{2x}$$

Square both sides of the equation.

$$\left(\sqrt{2x+33}\right)^{2}=\left(3+\sqrt{2x}\right)^{2}$$

Calculate \sqrt{2x+33} to the power of 2 and get 2x+33.

$$2x+33=\left(3+\sqrt{2x}\right)^{2}$$

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

$$2x+33=9+6\sqrt{2x}+\left(\sqrt{2x}\right)^{2}$$

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

$$2x+33=9+6\sqrt{2x}+2x$$

Subtract 6\sqrt{2x} from both sides.

$$2x+33-6\sqrt{2x}=9+2x$$

Subtract 2x from both sides.

$$2x+33-6\sqrt{2x}-2x=9$$

Combine 2x and -2x to get 0.

$$33-6\sqrt{2x}=9$$

Subtract 33 from both sides.

$$-6\sqrt{2x}=9-33$$

Subtract 33 from 9 to get -24.

$$-6\sqrt{2x}=-24$$

Divide both sides by -6.

$$\sqrt{2x}=\frac{-24}{-6}$$

Divide -24 by -6 to get 4.

$$\sqrt{2x}=4$$

Square both sides of the equation.

$$2x=16$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{16}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Substitute 8 for x in the equation \sqrt{2x+33}-\sqrt{2x}=3.

$$\sqrt{2\times 8+33}-\sqrt{2\times 8}=3$$

Simplify. The value x=8 satisfies the equation.

$$3=3$$

Equation \sqrt{2x+33}=\sqrt{2x}+3 has a unique solution.

$$x=8$$