Solve sqrt{3/2}+1/2

$$\sqrt{ \frac{ 3 }{ 2 } } + \frac{ 1 }{ 2 }$$

Evaluate

$$\frac{\sqrt{6}+1}{2}\approx 1.724744871$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{3}{2}} as the division of square roots \frac{\sqrt{3}}{\sqrt{2}}.

$$\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}$$

Rationalize the denominator of \frac{\sqrt{3}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

$$\frac{\sqrt{3}\sqrt{2}}{2}+\frac{1}{2}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{\sqrt{6}}{2}+\frac{1}{2}$$

Since \frac{\sqrt{6}}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{6}+1}{2}$$

Factor

$$\frac{\sqrt{6} + 1}{2} = 1.724744871391589$$