Solve {left(frac{sqrt{3}-1}{2}right)}^2=

$${ \left( \frac{ \sqrt{ 3 } -1 }{ 2 } \right) }^{ 2 } =$$

Evaluate

$$-\frac{\sqrt{3}}{2}+1\approx 0.133974596$$

Solution Steps

To raise \frac{\sqrt{3}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{3}-1\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$\frac{\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3-2\sqrt{3}+1}{2^{2}}$$

Add 3 and 1 to get 4.

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

Calculate 2 to the power of 2 and get 4.

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

Expand

$$-\frac{\sqrt{3}}{2} + 1 = 0.133974596$$

Solution Steps

To raise \frac{\sqrt{3}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{3}-1\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$\frac{\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3-2\sqrt{3}+1}{2^{2}}$$

Add 3 and 1 to get 4.

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

Calculate 2 to the power of 2 and get 4.

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