Solve frac{sqrt{3}}{9-sqrt{2}}

$$\frac { \sqrt { 3 } } { 9 - \sqrt { 2 } }$$

Evaluate

$$\frac{\sqrt{3}\left(\sqrt{2}+9\right)}{79}\approx 0.228328443$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}}{9-\sqrt{2}} by multiplying numerator and denominator by 9+\sqrt{2}.

$$\frac{\sqrt{3}\left(9+\sqrt{2}\right)}{\left(9-\sqrt{2}\right)\left(9+\sqrt{2}\right)}$$

Consider \left(9-\sqrt{2}\right)\left(9+\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}.

$$\frac{\sqrt{3}\left(9+\sqrt{2}\right)}{9^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 9. Square \sqrt{2}.

$$\frac{\sqrt{3}\left(9+\sqrt{2}\right)}{81-2}$$

Subtract 2 from 81 to get 79.

$$\frac{\sqrt{3}\left(9+\sqrt{2}\right)}{79}$$

Use the distributive property to multiply \sqrt{3} by 9+\sqrt{2}.

$$\frac{9\sqrt{3}+\sqrt{3}\sqrt{2}}{79}$$

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

$$\frac{9\sqrt{3}+\sqrt{6}}{79}$$