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

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

Evaluate

$$\frac{\sqrt{2}+\sqrt{6}-2\sqrt{3}-1}{5}\approx -0.120079662$$

Solution Steps

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

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

Consider \left(1+\sqrt{6}\right)\left(1-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 1. Square \sqrt{6}.

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

Subtract 6 from 1 to get -5.

$$\frac{\left(1-\sqrt{2}\right)\left(1-\sqrt{6}\right)}{-5}$$

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

$$\frac{1-\sqrt{6}-\sqrt{2}+\sqrt{2}\sqrt{6}}{-5}$$

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}.

$$\frac{1-\sqrt{6}-\sqrt{2}+\sqrt{2}\sqrt{2}\sqrt{3}}{-5}$$

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

$$\frac{1-\sqrt{6}-\sqrt{2}+2\sqrt{3}}{-5}$$

Multiply both numerator and denominator by -1.

$$\frac{-1+\sqrt{6}+\sqrt{2}-2\sqrt{3}}{5}$$