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

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}-2x+1} to the power of 2 and get x^{2}-2x+1.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$x^{2}-2x+1=81-18x+x^{2}$$

Add 18x to both sides.

$$x^{2}-2x+1+18x=81+x^{2}$$

Combine -2x and 18x to get 16x.

$$x^{2}+16x+1=81+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+16x+1-x^{2}=81$$

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

$$16x+1=81$$

Subtract 1 from both sides.

$$16x=81-1$$

Subtract 1 from 81 to get 80.

$$16x=80$$

Divide both sides by 16.

$$x=\frac{80}{16}$$

Divide 80 by 16 to get 5.

$$x=5$$

Substitute 5 for x in the equation \sqrt{x^{2}-2x+1}=9-x.

$$\sqrt{5^{2}-2\times 5+1}=9-5$$

Simplify. The value x=5 satisfies the equation.

$$4=4$$

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

$$x=5$$