Solve sqrt{3/2}+sqrt{1/8}=

$$\sqrt { \frac { 3 } { 2 } } + \sqrt { \frac { 1 } { 8 } } =$$

Evaluate

$$\frac{\sqrt{2}\left(2\sqrt{3}+1\right)}{4}\approx 1.578298262$$

Short Solution Steps

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

$$\frac{\sqrt{3}}{\sqrt{2}}+\sqrt{\frac{1}{8}}$$

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

$$\frac{\sqrt{3}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+\sqrt{\frac{1}{8}}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{3}\sqrt{2}}{2}+\sqrt{\frac{1}{8}}$$

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

$$\frac{\sqrt{6}}{2}+\sqrt{\frac{1}{8}}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$\frac{\sqrt{6}}{2}+\frac{\sqrt{1}}{\sqrt{8}}$$

Calculate the square root of 1 and get 1.

$$\frac{\sqrt{6}}{2}+\frac{1}{\sqrt{8}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 4 is 4. Multiply \frac{\sqrt{6}}{2} times \frac{2}{2}.

$$\frac{2\sqrt{6}}{4}+\frac{\sqrt{2}}{4}$$

Since \frac{2\sqrt{6}}{4} and \frac{\sqrt{2}}{4} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{6}+\sqrt{2}}{4}$$