Solve frac{frac{sqrt{3}}{2}+1}{frac{sqrt{3}}{2}-1}

$$\frac { \frac { \sqrt { 3 } } { 2 } + 1 } { \frac { \sqrt { 3 } } { 2 } - 1 }$$

Evaluate

$$-4\sqrt{3}-7\approx -13.92820323$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

$$\frac{\frac{\sqrt{3}}{2}+\frac{2}{2}}{\frac{\sqrt{3}}{2}-1}$$

Since \frac{\sqrt{3}}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\sqrt{3}+2}{2}}{\frac{\sqrt{3}}{2}-1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{\sqrt{3}}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

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

Divide \frac{\sqrt{3}+2}{2} by \frac{\sqrt{3}-2}{2} by multiplying \frac{\sqrt{3}+2}{2} by the reciprocal of \frac{\sqrt{3}-2}{2}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

Consider \left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

$$\frac{7+4\sqrt{3}}{-1}$$

Anything divided by -1 gives its opposite. To find the opposite of 7+4\sqrt{3}, find the opposite of each term.

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

Factor

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