Solve (sqrt{3}-sqrt{2})(sqrt{8}+sqrt{12})

$$( \sqrt { 3 } - \sqrt { 2 } ) ( \sqrt { 8 } + \sqrt { 12 } )$$

Evaluate

$$2$$

Solution Steps

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}.

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

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}.

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

Apply the distributive property by multiplying each term of \sqrt{3}-\sqrt{2} by each term of 2\sqrt{2}+2\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{2} is 2.

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

Multiply -2 and 2 to get -4.

$$2\sqrt{6}+6-4-2\sqrt{3}\sqrt{2}$$

Subtract 4 from 6 to get 2.

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

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

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

Combine 2\sqrt{6} and -2\sqrt{6} to get 0.

$$2$$

Factor

$$2$$