Solve (sqrt{2}+sqrt{6})(sqrt{6}-2sqrt{2})

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

Evaluate

$$2-2\sqrt{3}\approx -1.464101615$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{2}+\sqrt{6} by each term of \sqrt{6}-2\sqrt{2}.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

The square of \sqrt{2} is 2.

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{6} is 6.

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

Add -4 and 6 to get 2.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -2 and 2 to get -4.

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

Combine 2\sqrt{3} and -4\sqrt{3} to get -2\sqrt{3}.

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