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

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

Evaluate

$$2$$

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}.

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

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}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

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

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

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

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

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

The square of \sqrt{2} is 2.

$$6-2\times 2$$

Multiply -2 and 2 to get -4.

$$6-4$$

Subtract 4 from 6 to get 2.

$$2$$

Factor

$$2$$