Solve frac{3+sqrt{5}}{3-sqrt{5}}=

$$\frac{ 3+ \sqrt{ 5 } }{ 3- \sqrt{ 5 } } =$$

Evaluate

$$\frac{3\sqrt{5}+7}{2}\approx 6.854101966$$

Solution Steps

Rationalize the denominator of \frac{3+\sqrt{5}}{3-\sqrt{5}} by multiplying numerator and denominator by 3+\sqrt{5}.

$$\frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}$$

Consider \left(3-\sqrt{5}\right)\left(3+\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}.

$$\frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 3. Square \sqrt{5}.

$$\frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{9-5}$$

Subtract 5 from 9 to get 4.

$$\frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}$$

Multiply 3+\sqrt{5} and 3+\sqrt{5} to get \left(3+\sqrt{5}\right)^{2}.

$$\frac{\left(3+\sqrt{5}\right)^{2}}{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{5}\right)^{2}.

$$\frac{9+6\sqrt{5}+\left(\sqrt{5}\right)^{2}}{4}$$

The square of \sqrt{5} is 5.

$$\frac{9+6\sqrt{5}+5}{4}$$

Add 9 and 5 to get 14.

$$\frac{14+6\sqrt{5}}{4}$$