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

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

Solve for x

$$x=4$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$x^{2}+33=9+6x+x^{2}$$

Subtract 6x from both sides.

$$x^{2}+33-6x=9+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+33-6x-x^{2}=9$$

Combine x^{2} and -x^{2} to get 0.

$$33-6x=9$$

Subtract 33 from both sides.

$$-6x=9-33$$

Subtract 33 from 9 to get -24.

$$-6x=-24$$

Divide both sides by -6.

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

Divide -24 by -6 to get 4.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x^{2}+33}-x=3.

$$\sqrt{4^{2}+33}-4=3$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

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

$$x=4$$