Solve frac{sqrt{12}+sqrt{6}+sqrt{2}+2}{sqrt{3}+1}

$$\frac { \sqrt { 12 } + \sqrt { 6 } + \sqrt { 2 } + 2 } { \sqrt { 3 } + 1 }$$

Evaluate

$$\sqrt{2}+2\approx 3.414213562$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{2\sqrt{3}+\sqrt{6}+\sqrt{2}+2}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

$$\frac{\left(2\sqrt{3}+\sqrt{6}+\sqrt{2}+2\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}$$

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}.

$$\frac{\left(2\sqrt{3}+\sqrt{6}+\sqrt{2}+2\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}$$

Square \sqrt{3}. Square 1.

$$\frac{\left(2\sqrt{3}+\sqrt{6}+\sqrt{2}+2\right)\left(\sqrt{3}-1\right)}{3-1}$$

Subtract 1 from 3 to get 2.

$$\frac{\left(2\sqrt{3}+\sqrt{6}+\sqrt{2}+2\right)\left(\sqrt{3}-1\right)}{2}$$

Apply the distributive property by multiplying each term of 2\sqrt{3}+\sqrt{6}+\sqrt{2}+2 by each term of \sqrt{3}-1.

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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}.

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

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

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

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

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

Combine -\sqrt{6} and \sqrt{6} to get 0.

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

Combine 3\sqrt{2} and -\sqrt{2} to get 2\sqrt{2}.

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

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

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

Subtract 2 from 6 to get 4.

$$\frac{4+2\sqrt{2}}{2}$$

Divide each term of 4+2\sqrt{2} by 2 to get 2+\sqrt{2}.

$$2+\sqrt{2}$$