Solve frac{2-sqrt{3}}{2+sqrt{3}}=?

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

Evaluate

$$7-4\sqrt{3}\approx 0.07179677$$

Solution Steps

Rationalize the denominator of \frac{2-\sqrt{3}}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

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

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

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

Square 2. Square \sqrt{3}.

$$\frac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{4-3}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

$$4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

$$7-4\sqrt{3}$$