Solve sqrt{left(x-1right)^2+3^2}=x

$$\sqrt{ { \left(x-1 \right) }^{ 2 } + { 3 }^{ 2 } } =x$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(x-1\right)^{2}+3^{2}}\right)^{2}=x^{2}$$

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

$$\left(\sqrt{x^{2}-2x+1+3^{2}}\right)^{2}=x^{2}$$

Calculate 3 to the power of 2 and get 9.

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

Add 1 and 9 to get 10.

$$\left(\sqrt{x^{2}-2x+10}\right)^{2}=x^{2}$$

Calculate \sqrt{x^{2}-2x+10} to the power of 2 and get x^{2}-2x+10.

$$x^{2}-2x+10=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-2x+10-x^{2}=0$$

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

$$-2x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-2x=-10$$

Divide both sides by -2.

$$x=\frac{-10}{-2}$$

Divide -10 by -2 to get 5.

$$x=5$$

Substitute 5 for x in the equation \sqrt{\left(x-1\right)^{2}+3^{2}}=x.

$$\sqrt{\left(5-1\right)^{2}+3^{2}}=5$$

Simplify. The value x=5 satisfies the equation.

$$5=5$$

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

$$x=5$$