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

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

Evaluate

$$-3\sqrt{7}-8\approx -15.937253933$$

Solution Steps

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

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

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

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

Square \sqrt{7}. Square 3.

$$\frac{\left(\sqrt{7}+3\right)\left(\sqrt{7}+3\right)}{7-9}$$

Subtract 9 from 7 to get -2.

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

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

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

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

$$\frac{\left(\sqrt{7}\right)^{2}+6\sqrt{7}+9}{-2}$$

The square of \sqrt{7} is 7.

$$\frac{7+6\sqrt{7}+9}{-2}$$

Add 7 and 9 to get 16.

$$\frac{16+6\sqrt{7}}{-2}$$

Divide each term of 16+6\sqrt{7} by -2 to get -8-3\sqrt{7}.

$$-8-3\sqrt{7}$$