Solve (3/4sqrt{6}-sqrt{12})^2div1/2sqrt{3}

$$( \frac { 3 } { 4 } \sqrt { 6 } - \sqrt { 12 } ) ^ { 2 } \div \frac { 1 } { 2 } \sqrt { 3 }$$

Evaluate

$$\frac{123\sqrt{3}}{4}-18\sqrt{6}\approx 9.169746963$$

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}.

$$\frac{\left(\frac{3}{4}\sqrt{6}-2\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{4}\sqrt{6}-2\sqrt{3}\right)^{2}.

$$\frac{\frac{9}{16}\left(\sqrt{6}\right)^{2}-3\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

The square of \sqrt{6} is 6.

$$\frac{\frac{9}{16}\times 6-3\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

Multiply \frac{9}{16} and 6 to get \frac{27}{8}.

$$\frac{\frac{27}{8}-3\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{\frac{27}{8}-3\sqrt{3}\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{\frac{27}{8}-3\times 3\sqrt{2}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

Multiply -3 and 3 to get -9.

$$\frac{\frac{27}{8}-9\sqrt{2}+4\left(\sqrt{3}\right)^{2}}{\frac{1}{2}}\sqrt{3}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{27}{8}-9\sqrt{2}+4\times 3}{\frac{1}{2}}\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$\frac{\frac{27}{8}-9\sqrt{2}+12}{\frac{1}{2}}\sqrt{3}$$

Add \frac{27}{8} and 12 to get \frac{123}{8}.

$$\frac{\frac{123}{8}-9\sqrt{2}}{\frac{1}{2}}\sqrt{3}$$

Divide \frac{123}{8}-9\sqrt{2} by \frac{1}{2} by multiplying \frac{123}{8}-9\sqrt{2} by the reciprocal of \frac{1}{2}.

$$\left(\frac{123}{8}-9\sqrt{2}\right)\times 2\sqrt{3}$$

Use the distributive property to multiply \frac{123}{8}-9\sqrt{2} by 2.

$$\left(\frac{123}{4}-18\sqrt{2}\right)\sqrt{3}$$

Use the distributive property to multiply \frac{123}{4}-18\sqrt{2} by \sqrt{3}.

$$\frac{123}{4}\sqrt{3}-18\sqrt{2}\sqrt{3}$$

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

$$\frac{123}{4}\sqrt{3}-18\sqrt{6}$$