Solve (2*sqrt{3/2}-sqrt{1/2})*(1/2*sqrt{8}+sqrt{2/3})

$$(2 \times \sqrt{ \frac{ 3 }{ 2 } } - \sqrt{ \frac{ 1 }{ 2 } } ) \times ( \frac{ 1 }{ 2 } \times \sqrt{ 8 } + \sqrt{ \frac{ 2 }{ 3 } } )$$

Evaluate

$$\frac{5\sqrt{3}}{3}+1\approx 3.886751346$$

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}}.

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

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

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

The square of \sqrt{2} is 2.

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

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

$$\left(2\times \frac{\sqrt{6}}{2}-\sqrt{\frac{1}{2}}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

Cancel out 2 and 2.

$$\left(\sqrt{6}-\sqrt{\frac{1}{2}}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$\left(\sqrt{6}-\frac{\sqrt{1}}{\sqrt{2}}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

Calculate the square root of 1 and get 1.

$$\left(\sqrt{6}-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

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

$$\left(\sqrt{6}-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

The square of \sqrt{2} is 2.

$$\left(\sqrt{6}-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{2}{2}.

$$\left(\frac{2\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

Since \frac{2\sqrt{6}}{2} and \frac{\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\frac{1}{2}\sqrt{8}+\sqrt{\frac{2}{3}}\right)$$

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}.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\frac{1}{2}\times 2\sqrt{2}+\sqrt{\frac{2}{3}}\right)$$

Cancel out 2 and 2.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\sqrt{2}+\sqrt{\frac{2}{3}}\right)$$

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

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\sqrt{2}+\frac{\sqrt{2}}{\sqrt{3}}\right)$$

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

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\sqrt{2}+\frac{\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\right)$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\sqrt{2}+\frac{\sqrt{2}\sqrt{3}}{3}\right)$$

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

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\sqrt{2}+\frac{\sqrt{6}}{3}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{3}{3}.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\left(\frac{3\sqrt{2}}{3}+\frac{\sqrt{6}}{3}\right)$$

Since \frac{3\sqrt{2}}{3} and \frac{\sqrt{6}}{3} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{6}-\sqrt{2}}{2}\times \frac{3\sqrt{2}+\sqrt{6}}{3}$$

Multiply \frac{2\sqrt{6}-\sqrt{2}}{2} times \frac{3\sqrt{2}+\sqrt{6}}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2\sqrt{6}-\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{6}\right)}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{\left(2\sqrt{6}-\sqrt{2}\right)\left(3\sqrt{2}+\sqrt{6}\right)}{6}$$

Apply the distributive property by multiplying each term of 2\sqrt{6}-\sqrt{2} by each term of 3\sqrt{2}+\sqrt{6}.

$$\frac{6\sqrt{6}\sqrt{2}+2\left(\sqrt{6}\right)^{2}-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{6\sqrt{2}\sqrt{3}\sqrt{2}+2\left(\sqrt{6}\right)^{2}-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6\times 2\sqrt{3}+2\left(\sqrt{6}\right)^{2}-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

Multiply 6 and 2 to get 12.

$$\frac{12\sqrt{3}+2\left(\sqrt{6}\right)^{2}-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

The square of \sqrt{6} is 6.

$$\frac{12\sqrt{3}+2\times 6-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

Multiply 2 and 6 to get 12.

$$\frac{12\sqrt{3}+12-3\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{6}}{6}$$

The square of \sqrt{2} is 2.

$$\frac{12\sqrt{3}+12-3\times 2-\sqrt{2}\sqrt{6}}{6}$$

Multiply -3 and 2 to get -6.

$$\frac{12\sqrt{3}+12-6-\sqrt{2}\sqrt{6}}{6}$$

Subtract 6 from 12 to get 6.

$$\frac{12\sqrt{3}+6-\sqrt{2}\sqrt{6}}{6}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{12\sqrt{3}+6-\sqrt{2}\sqrt{2}\sqrt{3}}{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{12\sqrt{3}+6-2\sqrt{3}}{6}$$

Combine 12\sqrt{3} and -2\sqrt{3} to get 10\sqrt{3}.

$$\frac{10\sqrt{3}+6}{6}$$

Factor

$$\frac{\sqrt{3} {(\sqrt{3} + 5)}}{3} = 3.8867513459481287$$