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

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

Evaluate

$$-\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2}-\sqrt{3}-1\approx -4.66390246$$

Solution Steps

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

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

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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

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

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

Multiply both numerator and denominator by -1.

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

Factor

$$\frac{-\sqrt{2} - \sqrt{6} - 2 \sqrt{3} - 2}{2} = -4.663902460147014$$