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

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

Evaluate

$$-\sqrt{15}-4\approx -7.872983346$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square \sqrt{5}.

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

Subtract 5 from 3 to get -2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{3+2\sqrt{15}+\left(\sqrt{5}\right)^{2}}{-2}$$

The square of \sqrt{5} is 5.

$$\frac{3+2\sqrt{15}+5}{-2}$$

Add 3 and 5 to get 8.

$$\frac{8+2\sqrt{15}}{-2}$$

Divide each term of 8+2\sqrt{15} by -2 to get -4-\sqrt{15}.

$$-4-\sqrt{15}$$