Solve frac{-2-sqrt{3}}{-2+sqrt{3}}+frac{-2+sqrt{3}}{-2-sqrt{3}}

$$\frac{ -2- \sqrt{ 3 } }{ -2+ \sqrt{ 3 } } + \frac{ -2+ \sqrt{ 3 } }{ -2- \sqrt{ 3 } }$$

Evaluate

$$14$$

Solution Steps

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

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

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}.

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

Square -2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

Multiply -2-\sqrt{3} and -2-\sqrt{3} to get \left(-2-\sqrt{3}\right)^{2}.

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

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

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

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}.

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

Square -2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

Multiply -2+\sqrt{3} and -2+\sqrt{3} to get \left(-2+\sqrt{3}\right)^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Add 7 and 7 to get 14.

$$14+4\sqrt{3}-4\sqrt{3}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$14$$

Factor

$$2\times 7$$