Solve {l}{(7+4sqrt{3})(2-sqrt{3})^2+(2+sqrt{3})(2-sqrt{3})}{+sqrt{3}}

$$\left. \begin{array} { l } { ( 7 + 4 \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) ^ { 2 } + ( 2 + \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) } \\ { + \sqrt { 3 } } \end{array} \right.$$

Evaluate

$$\sqrt{3}+2\approx 3.732050808$$

Solution Steps

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

$$\left(7+4\sqrt{3}\right)\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

The square of \sqrt{3} is 3.

$$\left(7+4\sqrt{3}\right)\left(4-4\sqrt{3}+3\right)+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Add 4 and 3 to get 7.

$$\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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 7.

$$49-\left(4\sqrt{3}\right)^{2}+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$49-4^{2}\left(\sqrt{3}\right)^{2}+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Calculate 4 to the power of 2 and get 16.

$$49-16\left(\sqrt{3}\right)^{2}+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

The square of \sqrt{3} is 3.

$$49-16\times 3+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Multiply 16 and 3 to get 48.

$$49-48+\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\sqrt{3}$$

Subtract 48 from 49 to get 1.

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

Consider \left(2+\sqrt{3}\right)\left(2-\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 2.

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

The square of \sqrt{3} is 3.

$$1+4-3+\sqrt{3}$$

Subtract 3 from 4 to get 1.

$$1+1+\sqrt{3}$$

Add 1 and 1 to get 2.

$$2+\sqrt{3}$$