Solve (frac{sqrt{2}}{2}-sqrt{6})^2

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

Evaluate

$$\frac{13}{2}-2\sqrt{3}\approx 3.035898385$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{2}{2}.

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

Since \frac{\sqrt{2}}{2} and \frac{2\sqrt{6}}{2} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

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}.

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

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

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

Multiply -4 and 2 to get -8.

$$\frac{2-8\sqrt{3}+4\left(\sqrt{6}\right)^{2}}{2^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{2-8\sqrt{3}+4\times 6}{2^{2}}$$

Multiply 4 and 6 to get 24.

$$\frac{2-8\sqrt{3}+24}{2^{2}}$$

Add 2 and 24 to get 26.

$$\frac{26-8\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{26-8\sqrt{3}}{4}$$

Expand

$$\frac{13}{2} - 2 \sqrt{3} = 3.035898385$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{2}{2}.

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

Since \frac{\sqrt{2}}{2} and \frac{2\sqrt{6}}{2} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

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}.

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

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

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

Multiply -4 and 2 to get -8.

$$\frac{2-8\sqrt{3}+4\left(\sqrt{6}\right)^{2}}{2^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{2-8\sqrt{3}+4\times 6}{2^{2}}$$

Multiply 4 and 6 to get 24.

$$\frac{2-8\sqrt{3}+24}{2^{2}}$$

Add 2 and 24 to get 26.

$$\frac{26-8\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{26-8\sqrt{3}}{4}$$