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

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

Evaluate

$$\frac{\sqrt{21}+5}{2}\approx 4.791287847$$

Solution Steps

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

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

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

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

Square \sqrt{7}. Square \sqrt{3}.

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

Subtract 3 from 7 to get 4.

$$\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}+\sqrt{3}\right)}{4}$$

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

$$\frac{\left(\sqrt{3}+\sqrt{7}\right)^{2}}{4}$$

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

$$\frac{\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{4}$$

The square of \sqrt{3} is 3.

$$\frac{3+2\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{4}$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$\frac{3+2\sqrt{21}+\left(\sqrt{7}\right)^{2}}{4}$$

The square of \sqrt{7} is 7.

$$\frac{3+2\sqrt{21}+7}{4}$$

Add 3 and 7 to get 10.

$$\frac{10+2\sqrt{21}}{4}$$