Solve frac{(sqrt{6}+sqrt{2})^2}{(sqrt{7}+sqrt{3})(sqrt{7}-sqrt{3})}

$$\frac { ( \sqrt { 6 } + \sqrt { 2 } ) ^ { 2 } } { ( \sqrt { 7 } + \sqrt { 3 } ) ( \sqrt { 7 } - \sqrt { 3 } ) }$$

Evaluate

$$\sqrt{3}+2\approx 3.732050808$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+\sqrt{2}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

The square of \sqrt{6} is 6.

$$\frac{6+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

Multiply 2 and 2 to get 4.

$$\frac{6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

The square of \sqrt{2} is 2.

$$\frac{6+4\sqrt{3}+2}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

Add 6 and 2 to get 8.

$$\frac{8+4\sqrt{3}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}$$

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}.

$$\frac{8+4\sqrt{3}}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{8+4\sqrt{3}}{7-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{8+4\sqrt{3}}{7-3}$$

Subtract 3 from 7 to get 4.

$$\frac{8+4\sqrt{3}}{4}$$

Divide each term of 8+4\sqrt{3} by 4 to get 2+\sqrt{3}.

$$2+\sqrt{3}$$