Solve (sqrt{6}-sqrt{8})*(2sqrt{2}+sqrt{6})

$$( \sqrt { 6 } - \sqrt { 8 } ) \times ( 2 \sqrt { 2 } + \sqrt { 6 } )$$

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}.

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

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

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

The square of \sqrt{6} is 6.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$6-4\times 2$$

Multiply 4 and 2 to get 8.

$$6-8$$

Subtract 8 from 6 to get -2.

$$-2$$

Factor

$$-2$$