Solve sqrt{24}-sqrt{8}divsqrt{2/3}+(sqrt{3}+sqrt{2})(sqrt{3}-sqrt{2})

$$\sqrt { 24 } - \sqrt { 8 } \div \sqrt { \frac { 2 } { 3 } } + ( \sqrt { 3 } + \sqrt { 2 } ) ( \sqrt { 3 } - \sqrt { 2 } )$$

Evaluate

$$2\sqrt{6}+1-2\sqrt{3}\approx 2.43487787$$

Solution Steps

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}.

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

Rewrite the division of square roots \frac{\sqrt{8}}{\sqrt{\frac{2}{3}}} as the square root of the division \sqrt{\frac{8}{\frac{2}{3}}} and perform the division.

$$2\sqrt{6}-\sqrt{4}+\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

Calculate the square root of 4 and get 2.

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

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

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

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

Subtract 2 from 3 to get 1.

$$2\sqrt{6}-2+1$$

Add -2 and 1 to get -1.

$$2\sqrt{6}-1$$