Solve (2sqrt{2}-1)^2-(1+sqrt{3})*(sqrt{2}-sqrt{6})

$$( 2 \sqrt { 2 } - 1 ) ^ { 2 } - ( 1 + \sqrt { 3 } ) \times ( \sqrt { 2 } - \sqrt { 6 } )$$

Evaluate

$$9-2\sqrt{2}\approx 6.171572875$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-1\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-4\sqrt{2}+1-\left(1+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{6}\right)$$

The square of \sqrt{2} is 2.

$$4\times 2-4\sqrt{2}+1-\left(1+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{6}\right)$$

Multiply 4 and 2 to get 8.

$$8-4\sqrt{2}+1-\left(1+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{6}\right)$$

Add 8 and 1 to get 9.

$$9-4\sqrt{2}-\left(1+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{6}\right)$$

Use the distributive property to multiply 1+\sqrt{3} by \sqrt{2}-\sqrt{6}.

$$9-4\sqrt{2}-\left(\sqrt{2}-\sqrt{6}+\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{6}\right)$$

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

$$9-4\sqrt{2}-\left(\sqrt{2}-\sqrt{6}+\sqrt{6}-\sqrt{3}\sqrt{6}\right)$$

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

$$9-4\sqrt{2}-\left(\sqrt{2}-\sqrt{3}\sqrt{6}\right)$$

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

$$9-4\sqrt{2}-\left(\sqrt{2}-\sqrt{3}\sqrt{3}\sqrt{2}\right)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$9-4\sqrt{2}-\left(\sqrt{2}-3\sqrt{2}\right)$$

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

$$9-4\sqrt{2}-\left(-2\sqrt{2}\right)$$

The opposite of -2\sqrt{2} is 2\sqrt{2}.

$$9-4\sqrt{2}+2\sqrt{2}$$

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

$$9-2\sqrt{2}$$