Solve frac{sqrt{3}}{2sqrt{6}+2}

$$\frac { \sqrt { 3 } } { 2 \sqrt { 6 } + 2 }$$

Evaluate

$$\frac{\sqrt{3}\left(\sqrt{6}-1\right)}{10}\approx 0.251058988$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{6}+2} by multiplying numerator and denominator by 2\sqrt{6}-2.

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

Consider \left(2\sqrt{6}+2\right)\left(2\sqrt{6}-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}.

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

Expand \left(2\sqrt{6}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{3}\left(2\sqrt{6}-2\right)}{4\left(\sqrt{6}\right)^{2}-2^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{\sqrt{3}\left(2\sqrt{6}-2\right)}{4\times 6-2^{2}}$$

Multiply 4 and 6 to get 24.

$$\frac{\sqrt{3}\left(2\sqrt{6}-2\right)}{24-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{3}\left(2\sqrt{6}-2\right)}{24-4}$$

Subtract 4 from 24 to get 20.

$$\frac{\sqrt{3}\left(2\sqrt{6}-2\right)}{20}$$

Use the distributive property to multiply \sqrt{3} by 2\sqrt{6}-2.

$$\frac{2\sqrt{3}\sqrt{6}-2\sqrt{3}}{20}$$

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}.

$$\frac{2\sqrt{3}\sqrt{3}\sqrt{2}-2\sqrt{3}}{20}$$

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

$$\frac{2\times 3\sqrt{2}-2\sqrt{3}}{20}$$

Multiply 2 and 3 to get 6.

$$\frac{6\sqrt{2}-2\sqrt{3}}{20}$$