Solve 6/sqrt{12}+24/sqrt{12}-2sqrt{48}-3sqrt{8}

$$\frac { 6 } { \sqrt { 12 } } + \frac { 24 } { \sqrt { 12 } } - 2 \sqrt { 48 } - 3 \sqrt { 8 }$$

Evaluate

$$-3\sqrt{3}-6\sqrt{2}\approx -13.681433797$$

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}.

$$\frac{6}{2\sqrt{3}}+\frac{24}{\sqrt{12}}-2\sqrt{48}-3\sqrt{8}$$

Rationalize the denominator of \frac{6}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{6\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}+\frac{24}{\sqrt{12}}-2\sqrt{48}-3\sqrt{8}$$

The square of \sqrt{3} is 3.

$$\frac{6\sqrt{3}}{2\times 3}+\frac{24}{\sqrt{12}}-2\sqrt{48}-3\sqrt{8}$$

Cancel out 2\times 3 in both numerator and denominator.

$$\sqrt{3}+\frac{24}{\sqrt{12}}-2\sqrt{48}-3\sqrt{8}$$

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}.

$$\sqrt{3}+\frac{24}{2\sqrt{3}}-2\sqrt{48}-3\sqrt{8}$$

Rationalize the denominator of \frac{24}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{3}+\frac{24\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}-2\sqrt{48}-3\sqrt{8}$$

The square of \sqrt{3} is 3.

$$\sqrt{3}+\frac{24\sqrt{3}}{2\times 3}-2\sqrt{48}-3\sqrt{8}$$

Cancel out 2\times 3 in both numerator and denominator.

$$\sqrt{3}+4\sqrt{3}-2\sqrt{48}-3\sqrt{8}$$

Combine \sqrt{3} and 4\sqrt{3} to get 5\sqrt{3}.

$$5\sqrt{3}-2\sqrt{48}-3\sqrt{8}$$

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}.

$$5\sqrt{3}-2\times 4\sqrt{3}-3\sqrt{8}$$

Multiply -2 and 4 to get -8.

$$5\sqrt{3}-8\sqrt{3}-3\sqrt{8}$$

Combine 5\sqrt{3} and -8\sqrt{3} to get -3\sqrt{3}.

$$-3\sqrt{3}-3\sqrt{8}$$

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}.

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

Multiply -3 and 2 to get -6.

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