Solve frac{(sqrt{3}+sqrt{13})^2}{8+sqrt{39}}

$$\frac { ( \sqrt { 3 } + \sqrt { 13 } ) ^ { 2 } } { 8 + \sqrt { 39 } }$$

Evaluate

$$2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+\sqrt{13}\right)^{2}.

$$\frac{\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{13}+\left(\sqrt{13}\right)^{2}}{8+\sqrt{39}}$$

The square of \sqrt{3} is 3.

$$\frac{3+2\sqrt{3}\sqrt{13}+\left(\sqrt{13}\right)^{2}}{8+\sqrt{39}}$$

To multiply \sqrt{3} and \sqrt{13}, multiply the numbers under the square root.

$$\frac{3+2\sqrt{39}+\left(\sqrt{13}\right)^{2}}{8+\sqrt{39}}$$

The square of \sqrt{13} is 13.

$$\frac{3+2\sqrt{39}+13}{8+\sqrt{39}}$$

Add 3 and 13 to get 16.

$$\frac{16+2\sqrt{39}}{8+\sqrt{39}}$$

Rationalize the denominator of \frac{16+2\sqrt{39}}{8+\sqrt{39}} by multiplying numerator and denominator by 8-\sqrt{39}.

$$\frac{\left(16+2\sqrt{39}\right)\left(8-\sqrt{39}\right)}{\left(8+\sqrt{39}\right)\left(8-\sqrt{39}\right)}$$

Consider \left(8+\sqrt{39}\right)\left(8-\sqrt{39}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(16+2\sqrt{39}\right)\left(8-\sqrt{39}\right)}{8^{2}-\left(\sqrt{39}\right)^{2}}$$

Square 8. Square \sqrt{39}.

$$\frac{\left(16+2\sqrt{39}\right)\left(8-\sqrt{39}\right)}{64-39}$$

Subtract 39 from 64 to get 25.

$$\frac{\left(16+2\sqrt{39}\right)\left(8-\sqrt{39}\right)}{25}$$

Use the distributive property to multiply 16+2\sqrt{39} by 8-\sqrt{39} and combine like terms.

$$\frac{128-2\left(\sqrt{39}\right)^{2}}{25}$$

The square of \sqrt{39} is 39.

$$\frac{128-2\times 39}{25}$$

Multiply -2 and 39 to get -78.

$$\frac{128-78}{25}$$

Subtract 78 from 128 to get 50.

$$\frac{50}{25}$$

Divide 50 by 25 to get 2.

$$2$$

Factor

$$2$$