Solve frac{5+sqrt{11}}{5-sqrt{11}}

$$\frac { 5 + \sqrt { 11 } } { 5 - \sqrt { 11 } }$$

Evaluate

$$\frac{5\sqrt{11}+18}{7}\approx 4.940446279$$

Solution Steps

Rationalize the denominator of \frac{5+\sqrt{11}}{5-\sqrt{11}} by multiplying numerator and denominator by 5+\sqrt{11}.

$$\frac{\left(5+\sqrt{11}\right)\left(5+\sqrt{11}\right)}{\left(5-\sqrt{11}\right)\left(5+\sqrt{11}\right)}$$

Consider \left(5-\sqrt{11}\right)\left(5+\sqrt{11}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(5+\sqrt{11}\right)\left(5+\sqrt{11}\right)}{5^{2}-\left(\sqrt{11}\right)^{2}}$$

Square 5. Square \sqrt{11}.

$$\frac{\left(5+\sqrt{11}\right)\left(5+\sqrt{11}\right)}{25-11}$$

Subtract 11 from 25 to get 14.

$$\frac{\left(5+\sqrt{11}\right)\left(5+\sqrt{11}\right)}{14}$$

Multiply 5+\sqrt{11} and 5+\sqrt{11} to get \left(5+\sqrt{11}\right)^{2}.

$$\frac{\left(5+\sqrt{11}\right)^{2}}{14}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+\sqrt{11}\right)^{2}.

$$\frac{25+10\sqrt{11}+\left(\sqrt{11}\right)^{2}}{14}$$

The square of \sqrt{11} is 11.

$$\frac{25+10\sqrt{11}+11}{14}$$

Add 25 and 11 to get 36.

$$\frac{36+10\sqrt{11}}{14}$$