Solve (4)(sqrt{24}-sqrt{1/2})-(sqrt{1/8}+sqrt{6})

$$( 4 ) ( \sqrt { 24 } - \sqrt { \frac { 1 } { 2 } } ) - ( \sqrt { \frac { 1 } { 8 } } + \sqrt { 6 } )$$

Evaluate

$$-\frac{9\sqrt{2}}{4}+7\sqrt{6}\approx 13.964447684$$

Short Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$4\left(2\sqrt{6}-\sqrt{\frac{1}{2}}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

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

$$4\left(2\sqrt{6}-\frac{\sqrt{1}}{\sqrt{2}}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

Calculate the square root of 1 and get 1.

$$4\left(2\sqrt{6}-\frac{1}{\sqrt{2}}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

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

$$4\left(2\sqrt{6}-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

The square of \sqrt{2} is 2.

$$4\left(2\sqrt{6}-\frac{\sqrt{2}}{2}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

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

$$4\left(\frac{2\times 2\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

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

$$4\times \frac{2\times 2\sqrt{6}-\sqrt{2}}{2}-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

Do the multiplications in 2\times 2\sqrt{6}-\sqrt{2}.

$$4\times \frac{4\sqrt{6}-\sqrt{2}}{2}-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

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

$$2\left(4\sqrt{6}-\sqrt{2}\right)-\left(\sqrt{\frac{1}{8}}+\sqrt{6}\right)$$

Use the distributive property to multiply 2 by 4\sqrt{6}-\sqrt{2}.

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

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

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

Calculate the square root of 1 and get 1.

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$8\sqrt{6}-2\sqrt{2}-\left(\frac{\sqrt{2}}{4}+\sqrt{6}\right)$$

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

$$8\sqrt{6}-2\sqrt{2}-\left(\frac{\sqrt{2}}{4}+\frac{4\sqrt{6}}{4}\right)$$

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

$$8\sqrt{6}-2\sqrt{2}-\frac{\sqrt{2}+4\sqrt{6}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8\sqrt{6}-2\sqrt{2} times \frac{4}{4}.

$$\frac{4\left(8\sqrt{6}-2\sqrt{2}\right)}{4}-\frac{\sqrt{2}+4\sqrt{6}}{4}$$

Since \frac{4\left(8\sqrt{6}-2\sqrt{2}\right)}{4} and \frac{\sqrt{2}+4\sqrt{6}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\left(8\sqrt{6}-2\sqrt{2}\right)-\left(\sqrt{2}+4\sqrt{6}\right)}{4}$$

Do the multiplications in 4\left(8\sqrt{6}-2\sqrt{2}\right)-\left(\sqrt{2}+4\sqrt{6}\right).

$$\frac{32\sqrt{6}-8\sqrt{2}-\sqrt{2}-4\sqrt{6}}{4}$$

Do the calculations in 32\sqrt{6}-8\sqrt{2}-\sqrt{2}-4\sqrt{6}.

$$\frac{28\sqrt{6}-9\sqrt{2}}{4}$$