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

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

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

Solution Steps

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

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

Combine \sqrt{2} and -\sqrt{2} to get 0.

$$2\sqrt{3}$$