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

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

Evaluate

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

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

$$2-\sqrt{3}$$