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

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

Solve for x

$$x=3$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Add 6x to both sides.

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

Subtract x^{2} from both sides.

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

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

$$-9+6x=9$$

Add 9 to both sides.

$$6x=9+9$$

Add 9 and 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 \sqrt{x^{2}-9}-x=-3.

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

Simplify. The value x=3 satisfies the equation.

$$-3=-3$$

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

$$x=3$$