Solve sqrt{2/3}-1/6sqrt{24}-3/2sqrt{12}+left(sqrt{2}-sqrt{3}right)left(sqrt{2}+sqrt{3}right)

$$\sqrt{ \frac{ 2 }{ 3 } } - \frac{ 1 }{ 6 } \sqrt{ 24 } - \frac{ 3 }{ 2 } \sqrt{ 12 } + \left( \sqrt{ 2 } - \sqrt{ 3 } \right) \left( \sqrt{ 2 } + \sqrt{ 3 } \right)$$

Evaluate

$$-3\sqrt{3}-1\approx -6.196152423$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.

$$\frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

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

$$\frac{\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{2}\sqrt{3}}{3}-\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

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

$$\frac{\sqrt{6}}{3}-\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

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}.

$$\frac{\sqrt{6}}{3}-\frac{1}{6}\times 2\sqrt{6}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

Express -\frac{1}{6}\times 2 as a single fraction.

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

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

$$\frac{\sqrt{6}}{3}-\frac{1}{3}\sqrt{6}-\frac{3}{2}\sqrt{12}+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)$$

Combine \frac{\sqrt{6}}{3} and -\frac{1}{3}\sqrt{6} to get 0.

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

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}.

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

Cancel out 2 and 2.

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

Consider \left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{2} is 2.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 2 to get -1.

$$-3\sqrt{3}-1$$

Factor

$$-3\sqrt{3}-1$$