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

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

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

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}.

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

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

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

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

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

Combine -6\sqrt{3} and 2\sqrt{3} to get -4\sqrt{3}.

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

The square of \sqrt{3} is 3.

$$\frac{6-4\sqrt{3}-2\times 3}{-2}$$

Multiply -2 and 3 to get -6.

$$\frac{6-4\sqrt{3}-6}{-2}$$

Subtract 6 from 6 to get 0.

$$\frac{-4\sqrt{3}}{-2}$$

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

$$2\sqrt{3}$$