Solve frac{1+sqrt{15}}{1-sqrt{15}}

$$\frac{ 1+ \sqrt{ 15 } }{ 1- \sqrt{ 15 } }$$

Evaluate

$$\frac{-\sqrt{15}-8}{7}\approx -1.696140478$$

Solution Steps

Rationalize the denominator of \frac{1+\sqrt{15}}{1-\sqrt{15}} by multiplying numerator and denominator by 1+\sqrt{15}.

$$\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{\left(1-\sqrt{15}\right)\left(1+\sqrt{15}\right)}$$

Consider \left(1-\sqrt{15}\right)\left(1+\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}.

$$\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{1^{2}-\left(\sqrt{15}\right)^{2}}$$

Square 1. Square \sqrt{15}.

$$\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{1-15}$$

Subtract 15 from 1 to get -14.

$$\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{-14}$$

Multiply 1+\sqrt{15} and 1+\sqrt{15} to get \left(1+\sqrt{15}\right)^{2}.

$$\frac{\left(1+\sqrt{15}\right)^{2}}{-14}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{15}\right)^{2}.

$$\frac{1+2\sqrt{15}+\left(\sqrt{15}\right)^{2}}{-14}$$

The square of \sqrt{15} is 15.

$$\frac{1+2\sqrt{15}+15}{-14}$$

Add 1 and 15 to get 16.

$$\frac{16+2\sqrt{15}}{-14}$$