Solve frac{3+sqrt{15}}{3-sqrt{15}}

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

Evaluate

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

Solution Steps

Rationalize the denominator of \frac{3+\sqrt{15}}{3-\sqrt{15}} by multiplying numerator and denominator by 3+\sqrt{15}.

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

Consider \left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 3. Square \sqrt{15}.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{9-15}$$

Subtract 15 from 9 to get -6.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{-6}$$

Multiply 3+\sqrt{15} and 3+\sqrt{15} to get \left(3+\sqrt{15}\right)^{2}.

$$\frac{\left(3+\sqrt{15}\right)^{2}}{-6}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{15}\right)^{2}.

$$\frac{9+6\sqrt{15}+\left(\sqrt{15}\right)^{2}}{-6}$$

The square of \sqrt{15} is 15.

$$\frac{9+6\sqrt{15}+15}{-6}$$

Add 9 and 15 to get 24.

$$\frac{24+6\sqrt{15}}{-6}$$

Divide each term of 24+6\sqrt{15} by -6 to get -4-\sqrt{15}.

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