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

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

Solve for x

$$x=9$$

Solution Steps

Subtract -9 from both sides of the equation.

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

Add 9 and 9 to get 18.

$$\sqrt{x^{2}}=18-x$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}}\right)^{2}=\left(18-x\right)^{2}$$

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

$$x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$x^{2}=324-36x+x^{2}$$

Add 36x to both sides.

$$x^{2}+36x=324+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+36x-x^{2}=324$$

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

$$36x=324$$

Divide both sides by 36.

$$x=\frac{324}{36}$$

Divide 324 by 36 to get 9.

$$x=9$$

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

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

Simplify. The value x=9 satisfies the equation.

$$0=0$$

Equation \sqrt{x^{2}}=18-x has a unique solution.

$$x=9$$