Solve frac{sqrt{3}}{12+sqrt{6}}

$$\frac { \sqrt { 3 } } { 12 + \sqrt { 6 } }$$

Evaluate

$$\frac{2\sqrt{3}}{23}-\frac{\sqrt{2}}{46}\approx 0.119869341$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}}{12+\sqrt{6}} by multiplying numerator and denominator by 12-\sqrt{6}.

$$\frac{\sqrt{3}\left(12-\sqrt{6}\right)}{\left(12+\sqrt{6}\right)\left(12-\sqrt{6}\right)}$$

Consider \left(12+\sqrt{6}\right)\left(12-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{3}\left(12-\sqrt{6}\right)}{12^{2}-\left(\sqrt{6}\right)^{2}}$$

Square 12. Square \sqrt{6}.

$$\frac{\sqrt{3}\left(12-\sqrt{6}\right)}{144-6}$$

Subtract 6 from 144 to get 138.

$$\frac{\sqrt{3}\left(12-\sqrt{6}\right)}{138}$$

Use the distributive property to multiply \sqrt{3} by 12-\sqrt{6}.

$$\frac{12\sqrt{3}-\sqrt{3}\sqrt{6}}{138}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{12\sqrt{3}-\sqrt{3}\sqrt{3}\sqrt{2}}{138}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{12\sqrt{3}-3\sqrt{2}}{138}$$