Solve 6*(frac{sqrt{3}}{3})^2-sqrt{3}*frac{sqrt{3}}{2}-2*frac{sqrt{2}}{2}

$$6 \times ( \frac { \sqrt { 3 } } { 3 } ) ^ { 2 } - \sqrt { 3 } \times \frac { \sqrt { 3 } } { 2 } - 2 \times \frac { \sqrt { 2 } } { 2 }$$

Evaluate

$$\frac{1}{2}-\sqrt{2}\approx -0.914213562$$

Solution Steps

To raise \frac{\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

Express 6\times \frac{\left(\sqrt{3}\right)^{2}}{3^{2}} as a single fraction.

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

Express \sqrt{3}\times \frac{\sqrt{3}}{2} as a single fraction.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 2 is 18. Multiply \frac{6\left(\sqrt{3}\right)^{2}}{3^{2}} times \frac{2}{2}. Multiply \frac{3}{2} times \frac{9}{9}.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}}{18}-\frac{3\times 9}{18}-2\times \frac{\sqrt{2}}{2}$$

Since \frac{2\times 6\left(\sqrt{3}\right)^{2}}{18} and \frac{3\times 9}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-2\times \frac{\sqrt{2}}{2}$$

Cancel out 2 and 2.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{18}{18}.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\frac{18\sqrt{2}}{18}$$

Since \frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18} and \frac{18\sqrt{2}}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9-18\sqrt{2}}{18}$$

Do the multiplications.

$$\frac{12\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\sqrt{2}$$

The square of \sqrt{3} is 3.

$$\frac{12\times 3-3\times 9}{18}-\sqrt{2}$$

Multiply 12 and 3 to get 36.

$$\frac{36-3\times 9}{18}-\sqrt{2}$$

Multiply -3 and 9 to get -27.

$$\frac{36-27}{18}-\sqrt{2}$$

Subtract 27 from 36 to get 9.

$$\frac{9}{18}-\sqrt{2}$$

Reduce the fraction \frac{9}{18} to lowest terms by extracting and canceling out 9.

$$\frac{1}{2}-\sqrt{2}$$

Factor

$$\frac{1 - 2 \sqrt{2}}{2} = -0.9142135623730951$$