Solve sqrt{48}div(-sqrt{3})-sqrt{1/2}*sqrt{12}+sqrt{24}

$$\sqrt { 48 } \div ( - \sqrt { 3 } ) - \sqrt { \frac { 1 } { 2 } } \times \sqrt { 12 } + \sqrt { 24 }$$

Evaluate

$$\sqrt{6}-4\approx -1.550510257$$

Short Solution Steps

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}.

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\sqrt{\frac{1}{2}}\sqrt{12}+\sqrt{24}$$

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

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{\sqrt{1}}{\sqrt{2}}\sqrt{12}+\sqrt{24}$$

Calculate the square root of 1 and get 1.

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{1}{\sqrt{2}}\sqrt{12}+\sqrt{24}$$

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

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\sqrt{12}+\sqrt{24}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{\sqrt{2}}{2}\sqrt{12}+\sqrt{24}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{\sqrt{2}}{2}\times 2\sqrt{3}+\sqrt{24}$$

Cancel out 2 and 2.

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\sqrt{2}\sqrt{3}+\sqrt{24}$$

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

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\sqrt{6}+\sqrt{24}$$

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

$$\frac{4\sqrt{3}}{-\sqrt{3}}-\frac{\sqrt{6}\left(-\sqrt{3}\right)}{-\sqrt{3}}+\sqrt{24}$$

Since \frac{4\sqrt{3}}{-\sqrt{3}} and \frac{\sqrt{6}\left(-\sqrt{3}\right)}{-\sqrt{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\sqrt{3}-\sqrt{6}\left(-\sqrt{3}\right)}{-\sqrt{3}}+\sqrt{24}$$

Do the multiplications in 4\sqrt{3}-\sqrt{6}\left(-\sqrt{3}\right).

$$\frac{4\sqrt{3}+3\sqrt{2}}{-\sqrt{3}}+\sqrt{24}$$

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}.

$$\frac{4\sqrt{3}+3\sqrt{2}}{-\sqrt{3}}+2\sqrt{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2\sqrt{6} times \frac{-\sqrt{3}}{-\sqrt{3}}.

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

Since \frac{4\sqrt{3}+3\sqrt{2}}{-\sqrt{3}} and \frac{2\sqrt{6}\left(-\sqrt{3}\right)}{-\sqrt{3}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 4\sqrt{3}+3\sqrt{2}+2\sqrt{6}\left(-\sqrt{3}\right).

$$\frac{4\sqrt{3}+3\sqrt{2}-6\sqrt{2}}{-\sqrt{3}}$$

Do the calculations in 4\sqrt{3}+3\sqrt{2}-6\sqrt{2}.

$$\frac{4\sqrt{3}-3\sqrt{2}}{-\sqrt{3}}$$

Rationalize the denominator of \frac{4\sqrt{3}-3\sqrt{2}}{-\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

$$\frac{\left(4\sqrt{3}-3\sqrt{2}\right)\sqrt{3}}{-3}$$

Use the distributive property to multiply 4\sqrt{3}-3\sqrt{2} by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

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

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

Divide each term of 12-3\sqrt{6} by -3 to get -4+\sqrt{6}.

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