Solve frac{5+sqrt{3}}{6+sqrt{2}}

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

Evaluate

$$\frac{3\sqrt{3}}{17}-\frac{\sqrt{6}}{34}-\frac{5\sqrt{2}}{34}+\frac{15}{17}\approx 0.907992567$$

Solution Steps

Rationalize the denominator of \frac{5+\sqrt{3}}{6+\sqrt{2}} by multiplying numerator and denominator by 6-\sqrt{2}.

$$\frac{\left(5+\sqrt{3}\right)\left(6-\sqrt{2}\right)}{\left(6+\sqrt{2}\right)\left(6-\sqrt{2}\right)}$$

Consider \left(6+\sqrt{2}\right)\left(6-\sqrt{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}.

$$\frac{\left(5+\sqrt{3}\right)\left(6-\sqrt{2}\right)}{6^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 6. Square \sqrt{2}.

$$\frac{\left(5+\sqrt{3}\right)\left(6-\sqrt{2}\right)}{36-2}$$

Subtract 2 from 36 to get 34.

$$\frac{\left(5+\sqrt{3}\right)\left(6-\sqrt{2}\right)}{34}$$

Apply the distributive property by multiplying each term of 5+\sqrt{3} by each term of 6-\sqrt{2}.

$$\frac{30-5\sqrt{2}+6\sqrt{3}-\sqrt{3}\sqrt{2}}{34}$$

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

$$\frac{30-5\sqrt{2}+6\sqrt{3}-\sqrt{6}}{34}$$