Solve {l}{(4-sqrt{3})(4+sqrt{3})}{(1+sqrt{5})^2-sqrt{20}}

$$\left. \begin{array} { l } { ( 4 - \sqrt { 3 } ) ( 4 + \sqrt { 3 } ) } \\ { ( 1 + \sqrt { 5 } ) ^ { 2 } - \sqrt { 20 } } \end{array} \right.$$

Sort

$$6,13$$

Solution Steps

Consider \left(4-\sqrt{3}\right)\left(4+\sqrt{3}\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 4.

$$sort(16-\left(\sqrt{3}\right)^{2},\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

The square of \sqrt{3} is 3.

$$sort(16-3,\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

Subtract 3 from 16 to get 13.

$$sort(13,\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{5}\right)^{2}.

$$sort(13,1+2\sqrt{5}+\left(\sqrt{5}\right)^{2}-\sqrt{20})$$

The square of \sqrt{5} is 5.

$$sort(13,1+2\sqrt{5}+5-\sqrt{20})$$

Add 1 and 5 to get 6.

$$sort(13,6+2\sqrt{5}-\sqrt{20})$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$sort(13,6+2\sqrt{5}-2\sqrt{5})$$

Combine 2\sqrt{5} and -2\sqrt{5} to get 0.

$$sort(13,6)$$

To sort the list, start from a single element 13.

$$13$$

Insert 6 to the appropriate location in the new list.

$$6,13$$

Evaluate

$$13,\ 6$$