Solve (2sqrt{2}-sqrt{3})(sqrt{2}+2sqrt{3})-sqrt{24}

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

Evaluate

$$\sqrt{6}-2\approx 0.449489743$$

Solution Steps

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

$$2\left(\sqrt{2}\right)^{2}+4\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{2}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

The square of \sqrt{2} is 2.

$$2\times 2+4\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{2}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

Multiply 2 and 2 to get 4.

$$4+4\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{2}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

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

$$4+4\sqrt{6}-\sqrt{3}\sqrt{2}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

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

$$4+4\sqrt{6}-\sqrt{6}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

Combine 4\sqrt{6} and -\sqrt{6} to get 3\sqrt{6}.

$$4+3\sqrt{6}-2\left(\sqrt{3}\right)^{2}-\sqrt{24}$$

The square of \sqrt{3} is 3.

$$4+3\sqrt{6}-2\times 3-\sqrt{24}$$

Multiply -2 and 3 to get -6.

$$4+3\sqrt{6}-6-\sqrt{24}$$

Subtract 6 from 4 to get -2.

$$-2+3\sqrt{6}-\sqrt{24}$$

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}.

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

Combine 3\sqrt{6} and -2\sqrt{6} to get \sqrt{6}.

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