Solve sqrt{4+m^2}=m+1

$$\sqrt { 4 + m ^ { 2 } } = m + 1$$

Solve for m

$$m = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract m^{2} from both sides.

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

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

$$4=2m+1$$

Swap sides so that all variable terms are on the left hand side.

$$2m+1=4$$

Subtract 1 from both sides.

$$2m=4-1$$

Subtract 1 from 4 to get 3.

$$2m=3$$

Divide both sides by 2.

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

Substitute \frac{3}{2} for m in the equation \sqrt{4+m^{2}}=m+1.

$$\sqrt{4+\left(\frac{3}{2}\right)^{2}}=\frac{3}{2}+1$$

Simplify. The value m=\frac{3}{2} satisfies the equation.

$$\frac{5}{2}=\frac{5}{2}$$

Equation \sqrt{m^{2}+4}=m+1 has a unique solution.

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