Solve (sqrt{6}-sqrt{3/8})*sqrt{2}quad(2)(sqrt{32}+sqrt{2})

$$( \sqrt { 6 } - \sqrt { \frac { 3 } { 8 } } ) \times \sqrt { 2 } \quad ( 2 ) ( \sqrt { 32 } + \sqrt { 2 } )$$

Evaluate

$$15\sqrt{6}\approx 36.742346142$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{3}{8}} as the division of square roots \frac{\sqrt{3}}{\sqrt{8}}.

$$\left(\sqrt{6}-\frac{\sqrt{3}}{\sqrt{8}}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

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}.

$$\left(\sqrt{6}-\frac{\sqrt{3}}{2\sqrt{2}}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

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

$$\left(\sqrt{6}-\frac{\sqrt{3}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

The square of \sqrt{2} is 2.

$$\left(\sqrt{6}-\frac{\sqrt{3}\sqrt{2}}{2\times 2}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

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

$$\left(\sqrt{6}-\frac{\sqrt{6}}{2\times 2}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

Multiply 2 and 2 to get 4.

$$\left(\sqrt{6}-\frac{\sqrt{6}}{4}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{4}{4}.

$$\left(\frac{4\sqrt{6}}{4}-\frac{\sqrt{6}}{4}\right)\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

Since \frac{4\sqrt{6}}{4} and \frac{\sqrt{6}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\sqrt{6}-\sqrt{6}}{4}\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

Do the calculations in 4\sqrt{6}-\sqrt{6}.

$$\frac{3\sqrt{6}}{4}\sqrt{2}\times 2\left(\sqrt{32}+\sqrt{2}\right)$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{3\sqrt{6}}{4}\sqrt{2}\times 2\left(4\sqrt{2}+\sqrt{2}\right)$$

Combine 4\sqrt{2} and \sqrt{2} to get 5\sqrt{2}.

$$\frac{3\sqrt{6}}{4}\sqrt{2}\times 2\times 5\sqrt{2}$$

Multiply 2 and 5 to get 10.

$$\frac{3\sqrt{6}}{4}\sqrt{2}\times 10\sqrt{2}$$

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

$$\frac{3\sqrt{6}}{4}\times 2\times 10$$

Cancel out 4, the greatest common factor in 2 and 4.

$$\frac{3\sqrt{6}}{2}\times 10$$

Cancel out 2, the greatest common factor in 10 and 2.

$$5\times 3\sqrt{6}$$

Multiply 5 and 3 to get 15.

$$15\sqrt{6}$$