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

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

Evaluate

$$\frac{3\sqrt{6}}{2}+4\approx 7.674234614$$

Solution Steps

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

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

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

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

Square \sqrt{6}. Square 2.

$$\frac{\left(\sqrt{6}+1\right)\left(\sqrt{6}+2\right)}{6-4}$$

Subtract 4 from 6 to get 2.

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

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

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

The square of \sqrt{6} is 6.

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

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

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

Add 6 and 2 to get 8.

$$\frac{8+3\sqrt{6}}{2}$$