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

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

Evaluate

$$\frac{1}{2}=0.5$$

Short Solution Steps

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2^{2} and 2 is 4. Multiply \frac{\sqrt{3}+1}{2} times \frac{2}{2}.

$$\frac{\left(\sqrt{3}+1\right)^{2}}{4}-\frac{2\left(\sqrt{3}+1\right)}{4}$$

Since \frac{\left(\sqrt{3}+1\right)^{2}}{4} and \frac{2\left(\sqrt{3}+1\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(\sqrt{3}+1\right)^{2}-2\left(\sqrt{3}+1\right)}{4}$$

Do the multiplications in \left(\sqrt{3}+1\right)^{2}-2\left(\sqrt{3}+1\right).

$$\frac{\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1-2\sqrt{3}-2}{4}$$

Do the calculations in \left(\sqrt{3}\right)^{2}+2\sqrt{3}+1-2\sqrt{3}-2.

$$\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$