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

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

Evaluate

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

Solution Steps

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

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

Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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