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

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

Evaluate

$$\frac{\sqrt{2}}{4}+3\sqrt{6}\approx 7.702022619$$

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}.

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

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

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{2} is 2.

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

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

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

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

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

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

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

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

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

Calculate the square root of 1 and get 1.

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{12\sqrt{6}+\sqrt{2}}{4}$$