Solve frac{sqrt{-2}+1}{sqrt{-2}-1}

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

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{-2}+1}{\sqrt{-2}-1} by multiplying numerator and denominator by \sqrt{-2}+1.

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

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

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

Square \sqrt{-2}. Square 1.

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

Subtract 1 from -2 to get -3.

$$\frac{\left(\sqrt{-2}+1\right)\left(\sqrt{-2}+1\right)}{-3}$$

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

$$\frac{\left(\sqrt{-2}+1\right)^{2}}{-3}$$

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

$$\frac{\left(\sqrt{-2}\right)^{2}+2\sqrt{-2}+1}{-3}$$

Calculate \sqrt{-2} to the power of 2 and get -2.

$$\frac{-2+2\sqrt{-2}+1}{-3}$$

Add -2 and 1 to get -1.

$$\frac{-1+2\sqrt{-2}}{-3}$$

Multiply both numerator and denominator by -1.

$$\frac{1-2\sqrt{-2}}{3}$$