Solve (frac{4+2sqrt{3}}{2})^2+frac{4+2sqrt{3}}{2}*frac{4-2sqrt{3}}{2}-(frac{4-2sqrt{3}}{2})^2

$$( \frac { 4 + 2 \sqrt { 3 } } { 2 } ) ^ { 2 } + \frac { 4 + 2 \sqrt { 3 } } { 2 } \times \frac { 4 - 2 \sqrt { 3 } } { 2 } - ( \frac { 4 - 2 \sqrt { 3 } } { 2 } ) ^ { 2 }$$

Evaluate

$$8\sqrt{3}+1\approx 14.856406461$$

Solution Steps

Divide each term of 4+2\sqrt{3} by 2 to get 2+\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Divide each term of 4+2\sqrt{3} by 2 to get 2+\sqrt{3}.

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

Divide each term of 4-2\sqrt{3} by 2 to get 2-\sqrt{3}.

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

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.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 4 to get 1.

$$7+4\sqrt{3}+1-\left(\frac{4-2\sqrt{3}}{2}\right)^{2}$$

Add 7 and 1 to get 8.

$$8+4\sqrt{3}-\left(\frac{4-2\sqrt{3}}{2}\right)^{2}$$

Divide each term of 4-2\sqrt{3} by 2 to get 2-\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

$$8+4\sqrt{3}-\left(7-4\sqrt{3}\right)$$

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

$$8+4\sqrt{3}-7+4\sqrt{3}$$

Subtract 7 from 8 to get 1.

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

Combine 4\sqrt{3} and 4\sqrt{3} to get 8\sqrt{3}.

$$1+8\sqrt{3}$$