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

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

Evaluate

$$3-2\sqrt{3}\approx -0.464101615$$

Solution Steps

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

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

Consider \left(1+\sqrt{3}\right)\left(1-\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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

Apply the distributive property by multiplying each term of \sqrt{3}-3 by each term of 1-\sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from -3 to get -6.

$$\frac{\sqrt{3}-6+3\sqrt{3}}{-2}$$

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

$$\frac{4\sqrt{3}-6}{-2}$$

Divide each term of 4\sqrt{3}-6 by -2 to get -2\sqrt{3}+3.

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