Solve frac{7+sqrt{3}}{7-sqrt{3}}

$$\frac { 7 + \sqrt { 3 } } { 7 - \sqrt { 3 } }$$

Evaluate

$$\frac{7\sqrt{3}+26}{23}\approx 1.657580681$$

Solution Steps

Rationalize the denominator of \frac{7+\sqrt{3}}{7-\sqrt{3}} by multiplying numerator and denominator by 7+\sqrt{3}.

$$\frac{\left(7+\sqrt{3}\right)\left(7+\sqrt{3}\right)}{\left(7-\sqrt{3}\right)\left(7+\sqrt{3}\right)}$$

Consider \left(7-\sqrt{3}\right)\left(7+\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}.

$$\frac{\left(7+\sqrt{3}\right)\left(7+\sqrt{3}\right)}{7^{2}-\left(\sqrt{3}\right)^{2}}$$

Square 7. Square \sqrt{3}.

$$\frac{\left(7+\sqrt{3}\right)\left(7+\sqrt{3}\right)}{49-3}$$

Subtract 3 from 49 to get 46.

$$\frac{\left(7+\sqrt{3}\right)\left(7+\sqrt{3}\right)}{46}$$

Multiply 7+\sqrt{3} and 7+\sqrt{3} to get \left(7+\sqrt{3}\right)^{2}.

$$\frac{\left(7+\sqrt{3}\right)^{2}}{46}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+\sqrt{3}\right)^{2}.

$$\frac{49+14\sqrt{3}+\left(\sqrt{3}\right)^{2}}{46}$$

The square of \sqrt{3} is 3.

$$\frac{49+14\sqrt{3}+3}{46}$$

Add 49 and 3 to get 52.

$$\frac{52+14\sqrt{3}}{46}$$