Solve sqrt{(m-2)^2}=m+1

$$\sqrt { ( m - 2 ) ^ { 2 } } = m + 1$$

Solve for m

$$m=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

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

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

$$\left(\sqrt{m^{2}-4m+4}\right)^{2}=\left(m+1\right)^{2}$$

Calculate \sqrt{m^{2}-4m+4} to the power of 2 and get m^{2}-4m+4.

$$m^{2}-4m+4=\left(m+1\right)^{2}$$

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

$$m^{2}-4m+4=m^{2}+2m+1$$

Subtract m^{2} from both sides.

$$m^{2}-4m+4-m^{2}=2m+1$$

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

$$-4m+4=2m+1$$

Subtract 2m from both sides.

$$-4m+4-2m=1$$

Combine -4m and -2m to get -6m.

$$-6m+4=1$$

Subtract 4 from both sides.

$$-6m=1-4$$

Subtract 4 from 1 to get -3.

$$-6m=-3$$

Divide both sides by -6.

$$m=\frac{-3}{-6}$$

Reduce the fraction \frac{-3}{-6} to lowest terms by extracting and canceling out -3.

$$m=\frac{1}{2}$$

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

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

Simplify. The value m=\frac{1}{2} satisfies the equation.

$$\frac{3}{2}=\frac{3}{2}$$

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

$$m=\frac{1}{2}$$