Solve frac{5+sqrt{3}}{5-sqrt{3}}

$$\frac { 5 + \sqrt { 3 } } { 5 - \sqrt { 3 } }$$

Evaluate

$$\frac{5\sqrt{3}+14}{11}\approx 2.060023094$$

Solution Steps

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

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

Consider \left(5-\sqrt{3}\right)\left(5+\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}.

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

Square 5. Square \sqrt{3}.

$$\frac{\left(5+\sqrt{3}\right)\left(5+\sqrt{3}\right)}{25-3}$$

Subtract 3 from 25 to get 22.

$$\frac{\left(5+\sqrt{3}\right)\left(5+\sqrt{3}\right)}{22}$$

Multiply 5+\sqrt{3} and 5+\sqrt{3} to get \left(5+\sqrt{3}\right)^{2}.

$$\frac{\left(5+\sqrt{3}\right)^{2}}{22}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+\sqrt{3}\right)^{2}.

$$\frac{25+10\sqrt{3}+\left(\sqrt{3}\right)^{2}}{22}$$

The square of \sqrt{3} is 3.

$$\frac{25+10\sqrt{3}+3}{22}$$

Add 25 and 3 to get 28.

$$\frac{28+10\sqrt{3}}{22}$$