Solve frac{(sqrt{8}+sqrt{6})(sqrt{8}-sqrt{6})}{3-sqrt{7}}

$$\frac { ( \sqrt { 8 } + \sqrt { 6 } ) ( \sqrt { 8 } - \sqrt { 6 } ) } { 3 - \sqrt { 7 } }$$

Evaluate

$$\sqrt{7}+3\approx 5.645751311$$

Solution Steps

Consider \left(\sqrt{8}+\sqrt{6}\right)\left(\sqrt{8}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{8}\right)^{2}-\left(\sqrt{6}\right)^{2}}{3-\sqrt{7}}$$

The square of \sqrt{8} is 8.

$$\frac{8-\left(\sqrt{6}\right)^{2}}{3-\sqrt{7}}$$

The square of \sqrt{6} is 6.

$$\frac{8-6}{3-\sqrt{7}}$$

Subtract 6 from 8 to get 2.

$$\frac{2}{3-\sqrt{7}}$$

Rationalize the denominator of \frac{2}{3-\sqrt{7}} by multiplying numerator and denominator by 3+\sqrt{7}.

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

Consider \left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 3. Square \sqrt{7}.

$$\frac{2\left(3+\sqrt{7}\right)}{9-7}$$

Subtract 7 from 9 to get 2.

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

Cancel out 2 and 2.

$$3+\sqrt{7}$$