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

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

Evaluate

$$\sqrt{6}+3\sqrt{2}-2\sqrt{3}-2\approx 1.228028815$$

Solution Steps

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

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

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\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}.

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

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}.

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

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

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

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

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

To find the opposite of 2\sqrt{3}-3\sqrt{2}+2-\sqrt{6}, find the opposite of each term.

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

The opposite of -3\sqrt{2} is 3\sqrt{2}.

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

The opposite of -\sqrt{6} is \sqrt{6}.

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