Solve frac{1+sqrt{31}}{1-sqrt{31}}

$$\frac{ 1+ \sqrt{ 31 } }{ 1- \sqrt{ 31 } }$$

Evaluate

$$\frac{-\sqrt{31}-16}{15}\approx -1.437850958$$

Solution Steps

Rationalize the denominator of \frac{1+\sqrt{31}}{1-\sqrt{31}} by multiplying numerator and denominator by 1+\sqrt{31}.

$$\frac{\left(1+\sqrt{31}\right)\left(1+\sqrt{31}\right)}{\left(1-\sqrt{31}\right)\left(1+\sqrt{31}\right)}$$

Consider \left(1-\sqrt{31}\right)\left(1+\sqrt{31}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(1+\sqrt{31}\right)\left(1+\sqrt{31}\right)}{1^{2}-\left(\sqrt{31}\right)^{2}}$$

Square 1. Square \sqrt{31}.

$$\frac{\left(1+\sqrt{31}\right)\left(1+\sqrt{31}\right)}{1-31}$$

Subtract 31 from 1 to get -30.

$$\frac{\left(1+\sqrt{31}\right)\left(1+\sqrt{31}\right)}{-30}$$

Multiply 1+\sqrt{31} and 1+\sqrt{31} to get \left(1+\sqrt{31}\right)^{2}.

$$\frac{\left(1+\sqrt{31}\right)^{2}}{-30}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{31}\right)^{2}.

$$\frac{1+2\sqrt{31}+\left(\sqrt{31}\right)^{2}}{-30}$$

The square of \sqrt{31} is 31.

$$\frac{1+2\sqrt{31}+31}{-30}$$

Add 1 and 31 to get 32.

$$\frac{32+2\sqrt{31}}{-30}$$