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

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

Evaluate

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

Solution Steps

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

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Multiply 1+\sqrt{3} and \sqrt{3}+1 to get \left(1+\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

$$-\frac{4+2\sqrt{3}}{2}$$

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

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

To find the opposite of 2+\sqrt{3}, find the opposite of each term.

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