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

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

Evaluate

$$\sqrt{2}\approx 1.414213562$$

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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}.

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

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

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

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

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

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

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

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

$$\frac{2\sqrt{2}}{2}$$

Cancel out 2 and 2.

$$\sqrt{2}$$