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

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

Evaluate

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

Solution Steps

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

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

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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

The square of \sqrt{3} is 3.

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

Subtract 1 from -3 to get -4.

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

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

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

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

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