Solve frac{sqrt{3}-sqrt{2}}{13+sqrt{2}}=

$$\frac { \sqrt { 3 } - \sqrt { 2 } } { 13 + \sqrt { 2 } } =$$

Evaluate

$$\frac{13\sqrt{3}+2-\sqrt{6}-13\sqrt{2}}{167}\approx 0.022050266$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}-\sqrt{2}}{13+\sqrt{2}} by multiplying numerator and denominator by 13-\sqrt{2}.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(13-\sqrt{2}\right)}{\left(13+\sqrt{2}\right)\left(13-\sqrt{2}\right)}$$

Consider \left(13+\sqrt{2}\right)\left(13-\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}.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(13-\sqrt{2}\right)}{13^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 13. Square \sqrt{2}.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(13-\sqrt{2}\right)}{169-2}$$

Subtract 2 from 169 to get 167.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(13-\sqrt{2}\right)}{167}$$

Apply the distributive property by multiplying each term of \sqrt{3}-\sqrt{2} by each term of 13-\sqrt{2}.

$$\frac{13\sqrt{3}-\sqrt{3}\sqrt{2}-13\sqrt{2}+\left(\sqrt{2}\right)^{2}}{167}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{13\sqrt{3}-\sqrt{6}-13\sqrt{2}+\left(\sqrt{2}\right)^{2}}{167}$$

The square of \sqrt{2} is 2.

$$\frac{13\sqrt{3}-\sqrt{6}-13\sqrt{2}+2}{167}$$