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

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

Evaluate

$$2\sqrt{2}-3\approx -0.171572875$$

Solution Steps

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

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

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

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

Square 1. Square \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Add 1 and 2 to get 3.

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

Anything divided by -1 gives its opposite. To find the opposite of 3-2\sqrt{2}, find the opposite of each term.

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

The opposite of -2\sqrt{2} is 2\sqrt{2}.

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