Solve x=3+sqrt{x^2-9}

$$x = 3 + \sqrt { x ^ { 2 } - 9 }$$

Solve for x

$$x=3$$

Solution Steps

Subtract 3 from both sides of the equation.

$$x-3=\sqrt{x^{2}-9}$$

Square both sides of the equation.

$$\left(x-3\right)^{2}=\left(\sqrt{x^{2}-9}\right)^{2}$$

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

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

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

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

Subtract x^{2} from both sides.

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

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

$$-6x+9=-9$$

Subtract 9 from both sides.

$$-6x=-9-9$$

Subtract 9 from -9 to get -18.

$$-6x=-18$$

Divide both sides by -6.

$$x=\frac{-18}{-6}$$

Divide -18 by -6 to get 3.

$$x=3$$

Substitute 3 for x in the equation x=3+\sqrt{x^{2}-9}.

$$3=3+\sqrt{3^{2}-9}$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Equation x-3=\sqrt{x^{2}-9} has a unique solution.

$$x=3$$