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

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

Evaluate

$$\sqrt{3}-2\approx -0.267949192$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square 3.

$$\frac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-3\right)}{3-9}$$

Subtract 9 from 3 to get -6.

$$\frac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-3\right)}{-6}$$

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

$$\frac{\left(\sqrt{3}-3\right)^{2}}{-6}$$

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

$$\frac{\left(\sqrt{3}\right)^{2}-6\sqrt{3}+9}{-6}$$

The square of \sqrt{3} is 3.

$$\frac{3-6\sqrt{3}+9}{-6}$$

Add 3 and 9 to get 12.

$$\frac{12-6\sqrt{3}}{-6}$$

Divide each term of 12-6\sqrt{3} by -6 to get -2+\sqrt{3}.

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