Solve 1/sqrt{2+sqrt{3}}

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

Evaluate

$$\sqrt{3}-\sqrt{2}\approx 0.317837245$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{2}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{2}-\sqrt{3}.

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

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

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

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

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

The opposite of -\sqrt{3} is \sqrt{3}.

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