Solve sqrt{x^2+9}-1=x

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

Solve for x

$$x=4$$

Solution Steps

Subtract -1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$x^{2}+9=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x^{2}+9-x^{2}=2x+1$$

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

$$9=2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$2x+1=9$$

Subtract 1 from both sides.

$$2x=9-1$$

Subtract 1 from 9 to get 8.

$$2x=8$$

Divide both sides by 2.

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

Divide 8 by 2 to get 4.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x^{2}+9}-1=x.

$$\sqrt{4^{2}+9}-1=4$$

Simplify. The value x=4 satisfies the equation.

$$4=4$$

Equation \sqrt{x^{2}+9}=x+1 has a unique solution.

$$x=4$$