Solve sqrt{4/3}-sqrt{1/18}+sqrt{1/3}-7sqrt{1/48}

$$\sqrt{ \frac{ 4 }{ 3 } } - \sqrt{ \frac{ 1 }{ 18 } } + \sqrt{ \frac{ 1 }{ 3 } } -7 \sqrt{ \frac{ 1 }{ 48 } }$$

Evaluate

$$\frac{5\sqrt{3}}{12}-\frac{\sqrt{2}}{6}\approx 0.485985576$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{4}{3}} as the division of square roots \frac{\sqrt{4}}{\sqrt{3}}.

$$\frac{\sqrt{4}}{\sqrt{3}}-\sqrt{\frac{1}{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Calculate the square root of 4 and get 2.

$$\frac{2}{\sqrt{3}}-\sqrt{\frac{1}{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

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

$$\frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\sqrt{\frac{1}{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{3}}{3}-\sqrt{\frac{1}{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Rewrite the square root of the division \sqrt{\frac{1}{18}} as the division of square roots \frac{\sqrt{1}}{\sqrt{18}}.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{1}}{\sqrt{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Calculate the square root of 1 and get 1.

$$\frac{2\sqrt{3}}{3}-\frac{1}{\sqrt{18}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{2\sqrt{3}}{3}-\frac{1}{3\sqrt{2}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Rationalize the denominator of \frac{1}{3\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{3\left(\sqrt{2}\right)^{2}}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{3\times 2}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Multiply 3 and 2 to get 6.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{6}+\sqrt{\frac{1}{3}}-7\sqrt{\frac{1}{48}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{6}+\frac{\sqrt{1}}{\sqrt{3}}-7\sqrt{\frac{1}{48}}$$

Calculate the square root of 1 and get 1.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}-7\sqrt{\frac{1}{48}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-7\sqrt{\frac{1}{48}}$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}-7\sqrt{\frac{1}{48}}$$

Combine \frac{2\sqrt{3}}{3} and \frac{\sqrt{3}}{3} to get \sqrt{3}.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\sqrt{\frac{1}{48}}$$

Rewrite the square root of the division \sqrt{\frac{1}{48}} as the division of square roots \frac{\sqrt{1}}{\sqrt{48}}.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{\sqrt{1}}{\sqrt{48}}$$

Calculate the square root of 1 and get 1.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{1}{\sqrt{48}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{1}{4\sqrt{3}}$$

Rationalize the denominator of \frac{1}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{\sqrt{3}}{4\times 3}$$

Multiply 4 and 3 to get 12.

$$\sqrt{3}-\frac{\sqrt{2}}{6}-7\times \frac{\sqrt{3}}{12}$$

Express -7\times \frac{\sqrt{3}}{12} as a single fraction.

$$\sqrt{3}-\frac{\sqrt{2}}{6}+\frac{-7\sqrt{3}}{12}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{6}{6}.

$$\frac{6\sqrt{3}}{6}-\frac{\sqrt{2}}{6}+\frac{-7\sqrt{3}}{12}$$

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

$$\frac{6\sqrt{3}-\sqrt{2}}{6}+\frac{-7\sqrt{3}}{12}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 12 is 12. Multiply \frac{6\sqrt{3}-\sqrt{2}}{6} times \frac{2}{2}.

$$\frac{2\left(6\sqrt{3}-\sqrt{2}\right)}{12}+\frac{-7\sqrt{3}}{12}$$

Since \frac{2\left(6\sqrt{3}-\sqrt{2}\right)}{12} and \frac{-7\sqrt{3}}{12} have the same denominator, add them by adding their numerators.

$$\frac{2\left(6\sqrt{3}-\sqrt{2}\right)-7\sqrt{3}}{12}$$

Do the multiplications in 2\left(6\sqrt{3}-\sqrt{2}\right)-7\sqrt{3}.

$$\frac{12\sqrt{3}-2\sqrt{2}-7\sqrt{3}}{12}$$

Do the calculations in 12\sqrt{3}-2\sqrt{2}-7\sqrt{3}.

$$\frac{5\sqrt{3}-2\sqrt{2}}{12}$$