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

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

Evaluate

$$-8\sqrt{3}\approx -13.856406461$$

Solution Steps

Rationalize the denominator of \frac{1}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Rationalize the denominator of \frac{1}{2-\sqrt{3}} by multiplying numerator and denominator by 2+\sqrt{3}.

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

Subtract 7 from 7 to get 0.

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

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

$$-8\sqrt{3}$$