Solve sqrt{12}+6sqrt{1/24}-sqrt{48}

$$\sqrt { 12 } + 6 \sqrt { \frac { 1 } { 24 } } - \sqrt { 48 }$$

Evaluate

$$\frac{\sqrt{6}}{2}-2\sqrt{3}\approx -2.239356744$$

Short Solution Steps

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}.

$$2\sqrt{3}+6\sqrt{\frac{1}{24}}-\sqrt{48}$$

Rewrite the square root of the division \sqrt{\frac{1}{24}} as the division of square roots \frac{\sqrt{1}}{\sqrt{24}}.

$$2\sqrt{3}+6\times \frac{\sqrt{1}}{\sqrt{24}}-\sqrt{48}$$

Calculate the square root of 1 and get 1.

$$2\sqrt{3}+6\times \frac{1}{\sqrt{24}}-\sqrt{48}$$

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}.

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

Rationalize the denominator of \frac{1}{2\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

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

The square of \sqrt{6} is 6.

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

Multiply 2 and 6 to get 12.

$$2\sqrt{3}+6\times \frac{\sqrt{6}}{12}-\sqrt{48}$$

Cancel out 12, the greatest common factor in 6 and 12.

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

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}.

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

Combine 2\sqrt{3} and -4\sqrt{3} to get -2\sqrt{3}.

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

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

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

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

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

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

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