Solve frac{2+sqrt{5}}{2-sqrt{5}}

$$\frac { 2 + \sqrt { 5 } } { 2 - \sqrt { 5 } }$$

Evaluate

$$-4\sqrt{5}-9\approx -17.94427191$$

Solution Steps

Rationalize the denominator of \frac{2+\sqrt{5}}{2-\sqrt{5}} by multiplying numerator and denominator by 2+\sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}$$

Consider \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 2. Square \sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{4-5}$$

Subtract 5 from 4 to get -1.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{-1}$$

Multiply 2+\sqrt{5} and 2+\sqrt{5} to get \left(2+\sqrt{5}\right)^{2}.

$$\frac{\left(2+\sqrt{5}\right)^{2}}{-1}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{5}\right)^{2}.

$$\frac{4+4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}$$

The square of \sqrt{5} is 5.

$$\frac{4+4\sqrt{5}+5}{-1}$$

Add 4 and 5 to get 9.

$$\frac{9+4\sqrt{5}}{-1}$$

Anything divided by -1 gives its opposite. To find the opposite of 9+4\sqrt{5}, find the opposite of each term.

$$-9-4\sqrt{5}$$