Solve frac{sqrt{6}+sqrt{2}}{4}*frac{sqrt{6}-sqrt{2}}{4}

$$\frac { \sqrt { 6 } + \sqrt { 2 } } { 4 } \times \frac { \sqrt { 6 } - \sqrt { 2 } } { 4 }$$

Evaluate

$$\frac{1}{4}=0.25$$

Solution Steps

Multiply \frac{\sqrt{6}+\sqrt{2}}{4} times \frac{\sqrt{6}-\sqrt{2}}{4} by multiplying numerator times numerator and denominator times denominator.

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

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

$$\frac{\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}}{4\times 4}$$

The square of \sqrt{6} is 6.

$$\frac{6-\left(\sqrt{2}\right)^{2}}{4\times 4}$$

The square of \sqrt{2} is 2.

$$\frac{6-2}{4\times 4}$$

Subtract 2 from 6 to get 4.

$$\frac{4}{4\times 4}$$

Multiply 4 and 4 to get 16.

$$\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$\frac{1}{4}$$

Factor

$$\frac{1}{2 ^ {2}} = 0.25$$