Solve frac{sqrt{6}-sqrt{2}}{4}*frac{sqrt{6}-1sqrt{2}}{4}

$$\frac { \sqrt { 6 } - \sqrt { 2 } } { 4 } \times \frac { \sqrt { 6 } - 1 \sqrt { 2 } } { 4 }$$

Evaluate

$$-\frac{\sqrt{3}}{4}+\frac{1}{2}\approx 0.066987298$$

Solution Steps

Multiply \frac{\sqrt{6}-\sqrt{2}}{4} and \frac{\sqrt{6}-\sqrt{2}}{4} to get \left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^{2}.

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

To raise \frac{\sqrt{6}-\sqrt{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

$$\frac{6-4\sqrt{3}+2}{4^{2}}$$

Add 6 and 2 to get 8.

$$\frac{8-4\sqrt{3}}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{8-4\sqrt{3}}{16}$$

Factor

$$\frac{2 - \sqrt{3}}{4} = 0.0669872981077807$$