Solve frac{2frac{sqrt{6}*1sqrt{2}}{4}}{sqrt{3}+1}

$$\frac { 2 \frac { \sqrt { 6 } \cdot 1 \sqrt { 2 } } { 4 } } { \sqrt { 3 } + 1 }$$

Evaluate

$$\frac{3-\sqrt{3}}{2}\approx 0.633974596$$

Solution Steps

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

$$\frac{2\times \frac{1\sqrt{2}\sqrt{3}\sqrt{2}}{4}}{\sqrt{3}+1}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2\times \frac{1\times 2\sqrt{3}}{4}}{\sqrt{3}+1}$$

Multiply 1 and 2 to get 2.

$$\frac{2\times \frac{2\sqrt{3}}{4}}{\sqrt{3}+1}$$

Divide 2\sqrt{3} by 4 to get \frac{1}{2}\sqrt{3}.

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

Cancel out 2 and 2.

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

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

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Use the distributive property to multiply \sqrt{3} by \sqrt{3}-1.

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

The square of \sqrt{3} is 3.

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