Solve sqrt{2}(1-sqrt{3})+(sqrt{2}-1)(sqrt{2}+1)=

$$\sqrt{ 2 } (1- \sqrt{ 3 } )+( \sqrt{ 2 } -1)( \sqrt{ 2 } +1)=$$

Evaluate

$$\sqrt{2}+1-\sqrt{6}\approx -0.03527618$$

Solution Steps

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

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

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

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

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{2} is 2.

$$\sqrt{2}-\sqrt{6}+2-1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\sqrt{2}-\sqrt{6}+2-1$$

Subtract 1 from 2 to get 1.

$$\sqrt{2}-\sqrt{6}+1$$