Solve (sqrt{3}-1)^2-(3+sqrt{5})*(3-sqrt{5})

$$( \sqrt { 3 } - 1 ) ^ { 2 } - ( 3 + \sqrt { 5 } ) \times ( 3 - \sqrt { 5 } )$$

Evaluate

$$-2\sqrt{3}\approx -3.464101615$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1-\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)$$

The square of \sqrt{3} is 3.

$$3-2\sqrt{3}+1-\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)$$

Add 3 and 1 to get 4.

$$4-2\sqrt{3}-\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)$$

Consider \left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$4-2\sqrt{3}-\left(9-\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$4-2\sqrt{3}-\left(9-5\right)$$

Subtract 5 from 9 to get 4.

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

Subtract 4 from 4 to get 0.

$$-2\sqrt{3}$$